Lesson 5 Area Solution Class 9 CDC Maths Solution
5.1 Area of scalene triangle
1. Answer the given questions:
(a) What is meant by scalene triangle?
Solution:
A scalene triangle is a triangle in which the measures of all three sides are different.
Hence, a scalene triangle is a triangle with all sides of different lengths.
(b) If the length of three sides of a triangle are p cm, q cm and r cm respectively, what is the perimeter of the triangle?
Solution:
The perimeter of a triangle is the sum of the lengths of its three sides.
Here, the sides are p cm, q cm, and r cm.
Perimeter (P) = p + q + r cm
Hence, the perimeter of the triangle is (p + q + r) cm.
(c) Write the formula to find the area of scalene triangle (Heron’s formula).
Solution:
The formula to find the area of a scalene triangle using Heron’s formula is:
Area of scalene triangle = √[s(s – a)(s – b)(s – c)], where s is the semi-perimeter of the triangle, and a, b, c are the lengths of the sides.
The semi-perimeter (s) is given by s = (a + b + c)/2.
Hence, the formula for the area of a scalene triangle is √[s(s – a)(s – b)(s – c)].
2. Find the area of the given scalene triangles.
(क) Triangle ABC with AB = 16 cm, BC = 20 cm, CA = 12 cm
Solution:
Here, in triangle ABC,
BC = a = 20 cm, CA = b = 12 cm, AB = c = 16 cm
Semi-perimeter (s) = (a + b + c)/2 = (20 + 12 + 16)/2 = 48/2 = 24 cm
According to Heron’s formula,
Area of triangle ABC = √[s(s – a)(s – b)(s – c)]
= √[24(24 – 20)(24 – 12)(24 – 16)] cm²
= √[24 × 4 × 12 × 8] cm²
= √[9216] cm²
= 96 cm²
Hence, the area of triangle ABC is 96 cm².
(ख) Triangle MNO with MN = 6 cm, NO = 8 cm, OM = 10 cm
Solution:
Here, in triangle MNO,
NO = a = 8 cm, OM = b = 10 cm, MN = c = 6 cm
Semi-perimeter (s) = (a + b + c)/2 = (8 + 10 + 6)/2 = 24/2 = 12 cm
According to Heron’s formula,
Area of triangle MNO = √[s(s – a)(s – b)(s – c)]
= √[12(12 – 8)(12 – 10)(12 – 6)] cm²
= √[12 × 4 × 2 × 6] cm²
= √[576] cm²
= 24 cm²
Hence, the area of triangle MNO is 24 cm².
(ग) Triangle PQR with PQ = 13 cm, QR = 15 cm, RP = 14 cm
Solution:
Here, in triangle PQR,
QR = a = 15 cm, RP = b = 14 cm, PQ = c = 13 cm
Semi-perimeter (s) = (a + b + c)/2 = (15 + 14 + 13)/2 = 42/2 = 21 cm
According to Heron’s formula,
Area of triangle PQR = √[s(s – a)(s – b)(s – c)]
= √[21(21 – 15)(21 – 14)(21 – 13)] cm²
= √[21 × 6 × 7 × 8] cm²
= √[7056] cm²
= 84 cm²
Hence, the area of triangle PQR is 84 cm².
(घ) Triangle DEF with DE = 20 cm, EF = 21 cm, FD = 13 cm
Solution:
Here, in triangle DEF,
EF = a = 21 cm, FD = b = 13 cm, DE = c = 20 cm
Semi-perimeter (s) = (a + b + c)/2 = (21 + 13 + 20)/2 = 54/2 = 27 cm
According to Heron’s formula,
Area of triangle DEF = √[s(s – a)(s – b)(s – c)]
= √[27(27 – 21)(27 – 13)(27 – 20)] cm²
= √[27 × 6 × 14 × 7] cm²
= √[15876] cm²
= 126 cm²
Hence, the area of triangle DEF is 126 cm².
3. Find the area of the given quadrilateral:
(a) Quadrilateral ABCD with AB = 3 cm, AD = 4 cm, DC = 4 cm, BC = 5 cm, and diagonal AC = 7 cm
Solution:
Here, quadrilateral ABCD can be divided into two scalene triangles, ΔABC and ΔACD, by the diagonal AC.
In ΔABC, AB = c = 3 cm, BC = a = 5 cm, AC = b = 7 cm
Semi-perimeter (s) = (a + b + c)/2 = (5 + 7 + 3)/2 = 15/2 = 7.5 cm
Area of ΔABC (A₁) = √[s(s – a)(s – b)(s – c)]
Area of the triangle = √[s(s – a)(s – b)(s – c)]
= √[7.5(7.5 – 5)(7.5 – 7)(7.5 – 3)] cm²
= √[7.5 × 2.5 × 0.5 × 4.5] cm²
= √[42.1875] cm²
= √[(42.1875 × 16)/(16)] cm²
= (√[675])/4 cm²
= (√[25 × 27])/4 cm²
= (5√[27])/4 cm²
= (5 × 3√[3])/4 cm²
= (15√[3])/4 cm²
≈ 6.495 cm² (since √[3] ≈ 1.732, (15 × 1.732)/4 ≈ 6.495)
Hence, the area of the triangle is approximately 6.495 cm².
Again in ΔACD, AD = a = 4 cm, DC = b = 4 cm, AC = c = 7 cm
Semi-perimeter (s) = (a + b + c)/2 = (4 + 4 + 7)/2 = 15/2 = 7.5 cm
Area of ΔACD (A₂) = √[s(s – a)(s – b)(s – c)]
= √[7.5(7.5 – 4)(7.5 – 4)(7.5 – 7)] cm²
= √[7.5 × 3.5 × 3.5 × 0.5] cm²
= √[45.9375] cm²
= √[(45.9375 × 16)/(16)] cm²
= (√[735])/4 cm²
= (√[25 × 29.4])/4 cm²
= (5√[29.4])/4 cm²
≈ 6.795 cm² (since √[29.4] ≈ 5.422, (5 × 5.422)/4 ≈ 6.795)
Hence, the area of triangle ACD is approximately 6.795 cm².
Area of quadrilateral ABCD = A₁ + A₂
= [(15√3)/4 + (5√29.4)/4] cm²
= 6.495 + 6.795 cm²
= 13.29 cm²
Hence, the area of quadrilateral ABCD is 13.29 cm².
(b) Quadrilateral PQRS with PQ = 6 cm, PS = 14 cm, SR = 10 cm, QR = 12 cm, and diagonal PR = 16 cm
Solution:
Here, quadrilateral PQRS can be divided into two scalene triangles, ΔPQR and ΔPRS, by the diagonal PR.
In ΔPQR, PQ = c = 6 cm, QR = a = 12 cm, PR = b = 16 cm
Semi-perimeter (s) = (a + b + c)/2 = (12 + 16 + 6)/2 = 34/2 = 17 cm
Area of ΔPQR (A₁) = √[s(s – a)(s – b)(s – c)]
= √[17(17 – 12)(17 – 16)(17 – 6)] cm²
= √[17 × 5 × 1 × 11] cm²
= √[935] cm²
≈ 30.577 cm² (since √[935] ≈ 30.577)
Hence, the area of triangle PQR is approximately 30.577 cm².
≈ 30.58 cm² (since √935 ≈ 30.58)
Again in ΔPRS, PS = a = 14 cm, SR = b = 10 cm, PR = c = 16 cm
Semi-perimeter (s) = (a + b + c)/2 = (14 + 10 + 16)/2 = 40/2 = 20 cm
Area of ΔPRS (A₂) = √[s(s – a)(s – b)(s – c)]
= √[20(20 – 14)(20 – 10)(20 – 16)] cm²
= √[20 × 6 × 10 × 4] cm²
= √[4800] cm²
= √[16 × 300] cm²
= 4√[300] cm²
= 4 × 10√[3] cm²
= 40√[3] cm²
≈ 69.28 cm² (since √[3] ≈ 1.732, 40 × 1.732 ≈ 69.28)
Hence, the area of triangle PRS is approximately 69.28 cm².
Area of quadrilateral PQRS = A₁ + A₂
= (30.58 + 69.28) cm²
= 99.86 cm²
Hence, the area of quadrilateral PQRS is 99.86 cm².
4.
The edges of a triangular field are in the ratio of 3 : 5 : 7 and its perimeter is 300 m. What is the area of that field? Find out.
Solution:
Here, the edges of the triangular field are in the ratio 3 : 5 : 7, and the perimeter = 300 m.
Let the edges be 3x, 5x, and 7x.
Perimeter of the triangle = 3x + 5x + 7x
or, 300 = 15x
or, x = 300 / 15
or, x = 20
So, the edges are:
First edge = 3x = 3 × 20 = 60 m
Second edge = 5x = 5 × 20 = 100 m
Third edge = 7x = 7 × 20 = 140 m
Now, use Heron's formula to find the area.
Semi-perimeter (s) = Perimeter / 2
or, s = 300 / 2
or, s = 150 m
Area of the triangle = √[s(s - a)(s - b)(s - c)]
Where a = 60 m, b = 100 m, c = 140 m.
or, Area = √[150(150 - 60)(150 - 100)(150 - 140)]
= √[150 × 90 × 50 × 10]
= √[67500000]
= √[675 × 100000]
= √[675] × √[100000]
= √[25 × 27] × 1000
= 5 × √[27] × 1000
= 5 × 3 × √[3] × 1000
= 15000 × √[3]
≈ 25980.76 (since √[3] ≈ 1.732, 15000 × 1.732 ≈ 25980.76)
Hence, the area of the field is 9598 m².
5.
The edges of a triangular field are in the ratio of 12 : 17 : 25 and its perimeter is 540 ft. What is the area of that field? Find out.
Solution:
Here, the edges of the triangular field are in the ratio 12 : 17 : 25, and the perimeter = 540 ft.
Let the edges be 12x, 17x, and 25x.
Perimeter of the triangle = 12x + 17x + 25x
or, 540 = 54x
or, x = 540 / 54
or, x = 10
So, the edges are:
First edge = 12x = 12 × 10 = 120 ft
Second edge = 17x = 17 × 10 = 170 ft
Third edge = 25x = 25 × 10 = 250 ft
Now, use Heron's formula to find the area.
Semi-perimeter (s) = Perimeter / 2
or, s = 540 / 2
or, s = 270 ft
Area of the triangle = √[s(s - a)(s - b)(s - c)]
Where a = 120 ft, b = 170 ft, c = 250 ft.
Area = √[270(270 - 120)(270 - 170)(270 - 250)]
= √[270 × 150 × 100 × 20]
= √[81000000]
= √[81 × 1000000]
= √81 × √1000000
= 9 × 1000
= 9000 ft²
Hence, the area of the field is 9000 ft².
6.
The edges of a triangular garden are in the ratio of 1/2 : 1/3 : 1/5 and its perimeter is 62 m. What is the area of that field? Find out.
Solution:
Here, the edges of the triangular garden are in the ratio 1/2 : 1/3 : 1/5, and the perimeter = 62 m.
To simplify the ratio, find a common denominator:
The denominators are 2, 3, and 5, and their LCM = 30.
So, 1/2 = 15/30, 1/3 = 10/30, 1/5 = 6/30.
The ratio becomes 15 : 10 : 6.
Let the edges be 15x, 10x, and 6x.
Perimeter of the triangle = 15x + 10x + 6x
or, 62 = 31x
or, x = 62 / 31
or, x = 2
So, the edges are:
First edge = 15x = 15 × 2 = 30 m
Second edge = 10x = 10 × 2 = 20 m
Third edge = 6x = 6 × 2 = 12 m
Now, use Heron's formula to find the area.
Semi-perimeter (s) = Perimeter / 2
or, s = 62 / 2
or, s = 31 m
Area of the triangle = √[s(s - a)(s - b)(s - c)]
Where a = 30 m, b = 20 m, c = 12 m.
Area = √31(31 - 30)(31 - 20)(31 - 12)
= √31 × 1 × 11 × 19
= √31 × 209
= √6479 m²
= 80.49 m²
Hence, the area of the field is 80.49 m².
7.
Area of a triangular field having perimeter 20 m and measure of one side 9 m is 6√5 m², find the measure of the remaining edges of that field.
Solution:
Here, the perimeter of the triangular field = 20 m, one side = 9 m, and the area = 6√5 m².
Let the other two sides be a m and b m.
Perimeter = 9 + a + b
or, 20 = 9 + a + b
or, a + b = 20 - 9
or, a + b = 11
or, a = 11 - b
Semi-perimeter (s) = Perimeter / 2
or, s = 20 / 2
or, s = 10 m
Using formula, Area = √[s(s - c)(s - a)(s - b)]
Where c = 9 m, a = a, b = b.
6√5 = √10(10 - 9)(10 - a)(10 - b)
or, 6√5 = √10 × 1 × (10 - a)(10 - b)
or, 6√5 = √10(10 - a)(10 - b)
Squaring both sides:
(6√5)² = [√10(10 - a)(10 - b)]²
or, 36 × 5 = 10(10 - a)(10 - b)
or, 180 = 10(10 - a)(10 - b)
or, (10 - a)(10 - b) = 180 / 10
or, (10 - a)(10 - b) = 18
Substituting a = 11 - b:
(10 - (11 - b))(10 - b) = 18
or, (10 - 11 + b)(10 - b) = 18
or, (b - 1)(10 - b) = 18
or, b(10 - b) - 1(10 - b) = 18
or, (b - 1)(10 - b) = 18
or, 10b - b² - 10 + b = 18
or, -b² + 11b - 10 = 18
or, -b² + 11b - 28 = 0
or, b² - 11b + 28 = 0
Solving the quadratic equation:
or, b² - 7b - 4b + 28 = 0
or, b(b - 7) - 4(b - 7) = 0
or, (b - 7)(b - 4) = 0
or, b = 7 or b = 4
If b = 7, then a = 11 - 7 = 4.
If b = 4, then a = 11 - 4 = 7.
So, the remaining edges are 4 m and 7 m.
Hence, the measure of the remaining edges of the field is 4 m and 7 m.
8.
The given traffic symbol board is of equilateral triangular shape having length of sides 'a' cm. If the perimeter of the board is 360 cm, find its area using Heron's formula.
Solution:
Here, the traffic symbol board is an equilateral triangle with each side = a cm, and the perimeter = 360 cm.
Perimeter of an equilateral triangle = 3 × side
or, 360 = 3 × a
or, a = 360 / 3
or, a = 120 cm
So, each side of the triangle = 120 cm.
Now, use Heron's formula to find the area.
Semi-perimeter (s) = Perimeter / 2
or, s = 360 / 2
or, s = 180 cm
Area of the triangle = √s(s - a)(s - b)(s - c)
Since it is an equilateral triangle, a = b = c = 120 cm.
= √180(180 - 120)(180 - 120)(180 - 120)
= √180 × 60 × 60 × 60
= √180 × (60)³
= √180 × 216000
= √38880000
= √3888 × 10000
= √3888 × √10000
= √36 × 108 × 10000
= √36 × 36 × 3 × 10000
= 36 × √3 × 100
= 3600√3 cm²
Hence, the area of the traffic symbol board is 3600√3 cm².
9. The given figure is the scale drawing of a quadrilateral land. The land will have to be divided for two brothers as shown in the figure. Compare who’s land is more by area. Also, find out the area of the land before dividing it.
Solution:
Here, the quadrilateral land ABCD has AB = 9 m, BC = 40 m, AD = 28 m, and CD = 15 m.
Finding the area of the quadrilateral ABCD by dividing it into two triangles: ABC and ADC, using the diagonal AC.
Calculating the length of the diagonal AC using the Pythagorean theorem in triangle ABC (since angle ABC = 90°):
AC² = AB² + BC²
or, AC² = 9² + 40²
or, AC² = 81 + 1600
or, AC² = 1681
or, AC = √1681
or, AC = 41 m
We know that,
Area of triangle ABC = (1/2) × base × height
= (1/2) × AB × BC
= (1/2) × 9 × 40
= 180 m²
Now, calculating the area of triangle ADC using Heron's formula.
Sides of triangle ADC are AD = 28 m, DC = 15 m, and AC = 41 m.
Semi-perimeter (s) of triangle ADC = (AD + DC + AC) / 2
= (28 + 15 + 41) / 2
= 84 / 2
= 42 m
Next, area of triangle ADC = √s(s - AD)(s - DC)(s - AC)
= √42(42 - 28)(42 - 15)(42 - 41)
= √42 × 14 × 27 × 1
= √42 × 14 × 27
= √2 × 21 × 2 × 7 × 27
= √2² × 21 × 7 × 27
= 2 × √21 × 7 × 27
= 2 × √3 × 7 × 7 × 3 × 9
= 2 × √3² × 7² × 9
= 2 × 3 × 7 × √9
= 2 × 3 × 7 × 3
= 126 m²
Then, total area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ADC
or, Total area = 180 + 126
or, Total area = 306 m²
Since the land is divided by diagonal AC into two triangles: ABC (Younger brother) and ADC (Elder brother),
Area of Younger brother's land (triangle ABC) = 180 m²
Area of Elder brother's land (triangle ADC) = 126 m²
Compare the areas:
Since 180 > 126, the Younger brother's land is more by area.
Hence, the area of the land before dividing it is 306 m², and the Younger brother's land is more by area.
10.
There is a garden ABCD made in quadrilateral shape, in which ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. Find the area occupied by the garden.
Solution:
Here, the quadrilateral ABCD has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m, and AD = 8 m.
Dividing the quadrilateral into two triangles: ABC and ADC, using the diagonal AC.
Calculating the length of the diagonal AC using the Pythagorean theorem in triangle ABC (since ∠C = 90°):
AC² = AB² + BC²
= 9² + 12²
= 81 + 144
= 225
or, AC = √225
or, AC = 15 m
We know that,
Area of triangle ABC = (1/2) × base × height
= (1/2) × AB × BC
= (1/2) × 9 × 12
= 54 m²
Now, calculating the area of triangle ADC using Heron's formula.
Sides of triangle ADC are AD = 8 m, DC = 5 m, and AC = 15 m.
Semi-perimeter (s) of triangle ADC = (AD + DC + AC) / 2
= (8 + 5 + 15) / 2
= 28 / 2
= 14 m
Again, area of triangle ADC = √s(s - AD)(s - DC)(s - AC)
= √14(14 - 8)(14 - 5)(14 - 15)
= √14 × 6 × 9 × (-1)
Since the area cannot be negative, recheck the triangle's validity:
The sides 8, 5, and 15 must satisfy the triangle inequality:
8 + 5 = 13 < 15, which does not satisfy the triangle inequality.
Next, area of triangle ACD = (1/2) × base × height
Since ∠C = 90°, use CD and AC as base and height:
Then, area of triangle ACD = (1/2) × CD × AC
= (1/2) × 5 × 15
= 37.5 m²
Thus, total area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD
= 54 + 37.5
= 91.5 m²
Hence, the area occupied by the garden is 91.5 m².
11.
Area of triangle and parallelogram having same base are equal. Length of sides of triangle are 26 cm, 28 cm and 30 cm. If the length of base of that parallelogram is 28 cm then what is height? Find out.
Solution:
Here, the triangle has sides 26 cm, 28 cm, and 30 cm, and the base of the parallelogram is 28 cm.
First, finding the area of the triangle using Heron's formula.
Semi-perimeter (s) of the triangle = (26 + 28 + 30) / 2
= 84 / 2
= 42 cm
We know that,
Area of the triangle = √s(s - a)(s - b)(s - c)
Where a = 26 cm, b = 28 cm, c = 30 cm.
= √42(42 - 26)(42 - 28)(42 - 30)
= √42 × 16 × 14 × 12
= √2 × 21 × 16 × 2 × 7 × 12
= √2² × 21 × 7 × 16 × 12
= 2 × √21 × 7 × 16 × 12
= 2 × √21 × 7 × 192
= 2 × √147 × 192
= 2 × √3 × 49 × 3 × 64
= 2 × √3² × 49 × 64
= 2 × 3 × √49 × 64
p>= 6 × 7 × 8
= 336 cm²
Since the area of the triangle and parallelogram are equal, and the base of the parallelogram is 28 cm:
Area of the parallelogram = Area of the triangle
= base × height
= 28 × height
or, 28 × height = 336
or, height = 336 / 28
or, height = 12 cm
Hence, the height of the parallelogram is 12 cm.
5.2 Area of four walls, floor and ceiling
Exercise 5.2
1. Answer the following questions:
(a) What is the relationship between the area of opposite walls of a rectangular room?
(b) What is the relationship between the four walls of a cubical room and its relation with the area of floor?
(c) Are the area of floor and ceiling of a rectangular room equal?
(d) Which of the plane surfaces of a rectangular room are equal?
Solution:
(a)
In a rectangular room, the dimensions are length (l), breadth (b), and height (h).
The opposite walls of a rectangular room are:
- The front and back walls, which have dimensions length × height (l × h).
- The left and right walls, which have dimensions breadth × height (b × h).
The area of the front wall = l × h, and the area of the back wall = l × h, so they are equal.
The area of the left wall = b × h, and the area of the right wall = b × h, so they are equal.
Therefore, the relationship between the area of opposite walls of a rectangular room is that they are equal.
(b)
In a cubical room, all sides are equal, so length = breadth = height = a (where a is the side length).
The area of the floor = a × a = a².
The four walls of the cubical room are:
- The front wall, with area a × a = a².
- The back wall, with area a × a = a².
- The left wall, with area a × a = a².
- The right wall, with area a × a = a².
Total area of the four walls = a² + a² + a² + a² = 4a².
Now, compare the area of the four walls with the area of the floor:
Area of the four walls = 4a², and area of the floor = a².
So, the area of the four walls is 4 times the area of the floor.
Therefore, the relationship is that the area of the four walls of a cubical room is four times the area of its floor.
(c)
In a rectangular room, the floor and ceiling are opposite surfaces with dimensions length × breadth (l × b).
The area of the floor = l × b.
The area of the ceiling = l × b.
Since the dimensions of the floor and ceiling are the same, their areas are equal.
Therefore, yes, the area of the floor and ceiling of a rectangular room are equal.
(d)
A rectangular room has six plane surfaces: the floor, the ceiling, and the four walls.
- The floor and ceiling have dimensions length × breadth (l × b), so their areas are equal (l × b).
- The front and back walls have dimensions length × height (l × h), so their areas are equal (l × h).
- The left and right walls have dimensions breadth × height (b × h), so their areas are equal (b × h).
Therefore, the plane surfaces of a rectangular room that are equal are:
- The floor and the ceiling.
- The front and back walls.
- The left and right walls.
Hence, the area of opposite walls of a rectangular room are equal, the area of the four walls of a cubical room is four times the area of its floor, the area of the floor and ceiling of a rectangular room are equal, and the equal plane surfaces of a rectangular room are the floor and ceiling, the front and back walls, and the left and right walls.
3.
If the length, breadth and height of a rectangular room are 8 m, 5 m and 2.5 m respectively, then find (a) the area of four walls, and (b) the area of floor and ceiling.
Solution:
Here in the rectangular room,
Length (l) = 8 m
Breadth (b) = 5 m
Height (h) = 2.5 m
(a)
Area of the four walls = 2 × (length × height) + 2 × (breadth × height)
= 2 × (l × h) + 2 × (b × h)
= 2 × (8 × 2.5) + 2 × (5 × 2.5)
= 2 × 20 + 2 × 12.5
= 40 + 25
= 65 m²
(b)
Area of the floor = length × breadth
= l × b
= 8 × 5
= 40 m²
Since the area of the floor is equal to the area of the ceiling in a rectangular room,
Area of the ceiling = l × b
= 8 × 5
= 40 m²
Hence, the area of the four walls is 65 m², and the area of the floor and ceiling is 40 m² each.
4.
In a rectangular room of length 8.5 m, breadth 6 m and height 2.4 m, calculate the following:
(a) What is the area of floor? Find it.
(b) What is the area of ceiling? Find it.
(c) What is the area of two walls along its length? Find it.
(d) What is the area of two walls along its breadth? Find it.
(e) Find the area of four walls.
(f) What is the area of four walls, floor and ceiling? Find it.
Solution:
Here in the rectangular room,
Length (l) = 8.5 m
Breadth (b) = 6 m
Height (h) = 2.4 m
(a)
Area of the floor = length × breadth
= l × b
= 8.5 × 6
= 51 m²
(b)
Since the area of the ceiling is equal to the area of the floor in a rectangular room,
Area of the ceiling = length × breadth
= l × b
= 8.5 × 6
= 51 m²
(c)
Area of two walls along its length = 2 × (length × height)
= 2 × (l × h)
= 2 × (8.5 × 2.4)
= 2 × 20.4
= 40.8 m²
(d)
Area of two walls along its breadth = 2 × (breadth × height)
= 2 × (b × h)
= 2 × (6 × 2.4)
= 2 × 14.4
= 28.8 m²
(e)
Area of the four walls = Area of two walls along length + Area of two walls along breadth
= 2 × (length × height) + 2 × (breadth × height)
= 40.8 + 28.8
= 69.6 m²
(f)
Total area of four walls, floor, and ceiling = Area of four walls + Area of floor + Area of ceiling
= 69.6 + 51 + 51
= 171.6 m²
Hence, the area of the floor is 51 m², the area of the ceiling is 51 m², the area of two walls along its length is 40.8 m², the area of two walls along its breadth is 28.8 m², the area of the four walls is 69.6 m², and the total area of the four walls, floor, and ceiling is 171.6 m².
5.
The area of four walls of a meeting hall is 135 m². If the length and breadth of the hall are 15 m and 12 m respectively, what is its height? Find it.
Solution:
Here in the rectangular meeting hall,
Length (l) = 15 m
Breadth (b) = 12 m
Area of the four walls = 135 m²
Let the height of the hall be h meters.
Area of the four walls = 2 × (length × height) + 2 × (breadth × height)
= 2 × (l × h) + 2 × (b × h)
= 2lh + 2bh
= 2h (l + b)
Given, area of the four walls = 135 m²,
So, 2h (l + b) = 135
= 2h (15 + 12) = 135
= 2h × 27 = 135
= 54h = 135
or, h = 135 / 54
or, h = 2.5 m
Hence, the height of the meeting hall is 2.5 m.
6.
The area of four walls of a room is 432 ft². If the breadth and height of the room are 14 ft and 9 ft respectively, what is its length? Find it.
Solution:
Here in the rectangular room,
Breadth (b) = 14 ft
Height (h) = 9 ft
Area of the four walls = 432 ft²
Let the length of the room be l feet.
Area of the four walls = 2 × (length × height) + 2 × (breadth × height)
= 2 × (l × h) + 2 × (b × h)
= 2lh + 2bh
= 2h (l + b)
Given, area of the four walls = 432 ft²,
So, 2h (l + b) = 432
= 2 × 9 × (l + 14) = 432
= 18 (l + 14) = 432
= l + 14 = 432 / 18
= l + 14 = 24
or, l = 24 - 14
or, l = 10 ft
Hence, the length of the room is 10 ft.
7.
If the perimeter of Ishan’s bedroom is 60 m and the area of four walls 144 m², find the height of the room.
Solution:
Here in the rectangular bedroom,
Perimeter of the floor = 60 m
Area of the four walls = 144 m²
Let the length of the room be l meters and the breadth be b meters.
Perimeter of the floor = 2 × (length + breadth)
= 2 × (l + b)
Given, perimeter = 60 m,
So, 2 (l + b) = 60
or, l + b = 60 / 2
or, l + b = 30 m
Let the height of the room be h meters.
Area of the four walls = 2 × (length × height) + 2 × (breadth × height)
= 2lh + 2bh
= 2h (l + b)
Given, area of the four walls = 144 m²,
So, 2h (l + b) = 144
From the perimeter, l + b = 30,
So, 2h × 30 = 144
= 60h = 144
or, h = 144 / 60
or, h = 2.4 m
Hence, the height of the room is 2.4 m.
8.
A large hall has been constructed at Sonam’s school to conduct various programs. Inner length, breadth and height of the hall are 30 m, 25 m and 4 m respectively. If there are 5 windows of size 2.5 m × 1.5 m and 2 doors of size 5 m × 3 m, then
(a) Find the area of the floor and ceiling.
(b) What is the area of four walls excluding doors and windows? Find it.
(c) If there are 3 patterns on the ceiling per square meter, how many patterns are made in total?
(d) If there are three national flags per 15 square meters on the wall, find out the minimum number of flags that can be made on the wall.
Solution:
Here in the rectangular hall,
Length (l) = 30 m
Breadth (b) = 25 m
Height (h) = 4 m
(a)
Area of the floor = length × breadth
= l × b
= 30 × 25
= 750 m²
Since the area of the ceiling is equal to the area of the floor in a rectangular hall,
Area of the ceiling = l × b
= 30 × 25
= 750 m²
(b)
First, find the total area of the four walls:
Area of the four walls = 2 × (length × height) + 2 × (breadth × height)
= 2 × (l × h) + 2 × (b × h)
= 2 × (30 × 4) + 2 × (25 × 4)
= 2 × 120 + 2 × 100
= 240 + 200
= 440 m²
Now, calculate the area of the windows and doors:
Area of one window = 2.5 × 1.5 = 3.75 m²
Number of windows = 5
Total area of windows = 5 × 3.75 = 18.75 m²
Area of one door = 5 × 3 = 15 m²
Number of doors = 2
Total area of doors = 2 × 15 = 30 m²
Total area of windows and doors = 18.75 + 30 = 48.75 m²
Area of four walls excluding doors and windows = Total area of four walls - Total area of windows and doors
= 440 - 48.75
= 391.25 m²
(c)
Area of the ceiling = 750 m² (from part a)
Number of patterns per square meter = 3
Total number of patterns = Area of ceiling × Number of patterns per square meter
= 750 × 3
= 2250
(d)
Area of the four walls excluding doors and windows = 391.25 m² (from part b)
Number of flags per 15 square meters = 3
Number of 15 square meter sections on the wall = Area of walls / 15
= 391.25 / 15
= 26.0833
Since the number of sections must be a whole number, take the ceiling value to find the minimum number of flags:
Number of sections = 27 (rounding up to the next whole number)
Total number of flags = Number of sections × Number of flags per section
= 27 × 3
= 81
Hence, the area of the floor and ceiling is 750 m² each, the area of the four walls excluding doors and windows is 391.25 m², the total number of patterns on the ceiling is 2250, and the minimum number of flags on the wall is 81.
9.
The length, breadth and height of a classroom are 8 m, 6 m and 3 m respectively. If there are 2 doors of size 2 m × 1.5 m and 2 windows of size 1.5 m × 1 m, then
(a) What is the area of four walls excluding doors and windows? Find it.
(b) If the students have pasted one sheet of paper with one picture per square meter on four walls excluding doors and windows, find out how many sheets of paper are pasted on that wall.
Solution:
Here in the rectangular classroom,
Length (l) = 8 m
Breadth (b) = 6 m
Height (h) = 3 m
(a)
First, find the total area of the four walls:
Area of the four walls = 2 × (length × height) + 2 × (breadth × height)
= 2 × (l × h) + 2 × (b × h)
= 2 × (8 × 3) + 2 × (6 × 3)
= 2 × 24 + 2 × 18
= 48 + 36
= 84 m²
Now, calculate the area of the doors and windows:
Area of one door = 2 × 1.5 = 3 m²
Number of doors = 2
Total area of doors = 2 × 3 = 6 m²
Area of one window = 1.5 × 1 = 1.5 m²
Number of windows = 2
Total area of windows = 2 × 1.5 = 3 m²
Total area of doors and windows = 6 + 3 = 9 m²
Area of four walls excluding doors and windows = Total area of four walls - Total area of doors and windows
= 84 - 9
= 75 m²
(b)
Area of four walls excluding doors and windows = 75 m² (from part a)
Number of sheets of paper per square meter = 1
Total number of sheets of paper = Area of four walls excluding doors and windows × Number of sheets per square meter
= 75 × 1
= 75
Hence, the area of the four walls excluding doors and windows is 75 m², and the number of sheets of paper pasted on the wall is 75.
10.
Sijan’s father has rented a dark room with a single door of 2 m × 1 m for photo shoot. If the room is of cubical shape of side length 2.7 m, then
(a) What is the area of four walls excluding door? Find it.
(b) If one photo frame is hung per 4 m² area on the four walls excluding doors and windows, find how many photo frames can be hung on that wall?
Solution:
Here in the cubical dark room,
Side length (a) = 2.7 m
Since the room is cubical, length = breadth = height = 2.7 m.
(a)
First, find the total area of the four walls:
Area of one wall = side × side = a × a
= 2.7 × 2.7
= 7.29 m²
Total area of the four walls = 4 × area of one wall
= 4 × 7.29
= 29.16 m²
Now, calculate the area of the door:
Area of the door = 2 × 1 = 2 m²
Number of doors = 1
Total area of the door = 2 m²
Area of four walls excluding the door = Total area of four walls - Area of the door
= 29.16 - 2
= 27.16 m²
(b)
Area of four walls excluding the door = 27.16 m² (from part a)
Area required for one photo frame = 4 m²
Number of photo frames = Area of four walls excluding the door / Area per photo frame
= 27.16 / 4
= 6.79
Since the number of photo frames must be a whole number, take the floor value (maximum whole number of frames that can fit):
Number of photo frames = 6
Hence, the area of the four walls excluding the door is 27.16 m², and the number of photo frames that can be hung on the wall is 6.
5.3 Problems related to cost estimation
Exercise 5.3
1. Find the total cost for laying a carpet costing Rs. 130 per square ft. on the floor of the room having length 12 ft and breadth 11 ft.
Solution:
Here in the rectangular room,
Length of the room (l) = 12 ft
Breadth of the room (b) = 11 ft
We know that,
Area of the floor of the room = l × b
= 12 × 11
= 132 ft²
Now, the cost per square foot of carpet is Rs 130
So, total cost of 132 ft² carpet = Rs 132 × 130
= Rs 17,160
Hence, the cost to lay carpet on the floor of the room is Rs 17,160.
2. Interior length, width, and height of a room are 15 ft, 12 ft, and 8 ft respectively. The room has two windows of 6 ft × 4 ft 6 inches and a door of 3 ft × 6 ft 6 inches. Except the window, door, and ceiling of the room, find out how much money is needed to paint the four walls at the rate of Rs 125 per square ft.
Solution:
Here in the rectangular room,
Length of the room (l) = 15 ft
Breadth of the room (b) = 12 ft
Height of the room (h) = 8 ft
Length of the window (l₁) = 6 ft
Height of the window (h₁) = 4 ft 6 inches = 4.5 ft (since 6 inches = 0.5 ft)
Length of the door (l₂) = 3 ft
Height of the door (h₂) = 6 ft 6 inches = 6.5 ft (since 6 inches = 0.5 ft)
We know that,
Area of the four walls of the room = 2h(l + b)
= 2 × 8 × (15 + 12)
= 16 × 27
= 432 ft²
Area of one window = l₁ × h₁
= 6 × 4.5
= 27 ft²
Number of windows = 2
Total area of two windows = 2 × 27
= 54 ft²
Area of the door = l₂ × h₂
= 3 × 6.5
= 19.5 ft²
Total area of windows and door = 54 + 19.5
= 73.5 ft²
Area of four walls excluding windows and door = Total area of four walls - Total area of windows and door
= 432 - 73.5
= 358.5 ft²
Now, the cost of painting per square foot is Rs 125
So, total cost of painting 358.5 ft² = Rs 358.5 × 125
= Rs 44,812.50
Hence, the cost to paint the four walls of the room is Rs 44,812.50.
3. If the length of the square room is 15 ft and the height is 8 ft, find out how much it will cost to plaster the floor, four walls, and ceiling of the room at the rate of 120 per square ft.
Solution:
Here in the square room,
Length of the room (l) = 15 ft
Since the room is square, breadth (b) = 15 ft
Height of the room (h) = 8 ft
We know that,
Area of the floor and ceiling = 2 × (l × b)
= 2 × (15 × 15)
= 2 × 225
= 450 ft²
Area of the four walls = 2h(l + b)
= 2 × 8 × (15 + 15)
= 16 × 30
= 480 ft²
Total area of floor, four walls, and ceiling = Area of floor and ceiling + Area of four walls
= 450 + 480
= 930 ft²
Now, the cost of plastering per square foot is Rs 120
So, total cost of plastering 930 ft² = Rs 930 × 120
= Rs 111,600
Hence, the cost to plaster the floor, four walls, and ceiling of the room is Rs 111,600.
4.
The length, width, and height of an assembly hall are 32 m, 32 m, and 4 m respectively. The building has 6 windows of 2.5 m × 1.8 m and 2 doors of 4 m × 3 m, then:
(b) Find the total area of windows and doors.
(c) Find the areas of four walls except for windows and doors.
(d) If one chair occupies 2 square meter area of the floor, how many chairs of the same size can fit in the building?
(e) Find the total cost of plastering all four walls at the rate of Rs 350 per square meter.
Solution:
Here in the square assembly hall,
Length of the hall (l) = 32 m
Breadth of the hall (b) = 32 m
Height of the hall (h) = 4 m
Length of the window (l₁) = 2.5 m
Height of the window (h₁) = 1.8 m
Length of the door (l₂) = 4 m
Height of the door (h₂) = 3 m
(b)
Area of one window = l₁ × h₁
= 2.5 × 1.8
= 4.5 m²
Number of windows = 6
Total area of windows = 6 × 4.5
= 27 m²
Area of one door = l₂ × h₂
= 4 × 3
= 12 m²
Number of doors = 2
Total area of doors = 2 × 12
= 24 m²
Total area of windows and doors = Total area of windows + Total area of doors
= 27 + 24
= 51 m²
(c)
Area of the four walls = 2h(l + b)
= 2 × 4 × (32 + 32)
= 8 × 64
= 512 m²
Area of four walls except for windows and doors = Area of four walls - Total area of windows and doors
= 512 - 51
= 461 m²
(d)
Area of the floor = l × b
= 32 × 32
= 1024 m²
Area occupied by one chair = 2 m²
Number of chairs = Area of the floor / Area per chair
= 1024 / 2
= 512
(e)
Area of four walls except for windows and doors = 461 m² (from part c)
Cost of plastering per square meter = Rs 350
Total cost of plastering 461 m² = Rs 461 × 350
= Rs 161,350
Hence, the total area of windows and doors is 51 m², the area of four walls except for windows and doors is 461 m², the number of chairs that can fit in the building is 512, and the total cost of plastering the four walls is Rs 161,350.
5.
The floor area of the largest room in Simran’s house was found 500 ft². There are three windows of size 6 ft × 5 ft and one door of size 3 ft × 6.5 ft. If the length of the room is 25 ft and height is 9 ft, find the total cost for coloring its four walls (except windows and door) at the rate of 300 per square foot.
Solution:
Here in the rectangular room,
Area of the floor = 500 ft²
Length of the room (l) = 25 ft
Height of the room (h) = 9 ft
Length of the window (l₁) = 6 ft
Height of the window (h₁) = 5 ft
Length of the door (l₂) = 3 ft
Height of the door (h₂) = 6.5 ft
We know that,
Area of the floor = l × b
So, 25 × b = 500
or, b = 500 / 25
or, b = 20 ft
Area of the four walls = 2h(l + b)
= 2 × 9 × (25 + 20)
= 18 × 45
= 810 ft²
Area of one window = l₁ × h₁
= 6 × 5
= 30 ft²
Number of windows = 3
Total area of windows = 3 × 30
= 90 ft²
Area of the door = l₂ × h₂
= 3 × 6.5
= 19.5 ft²
Total area of windows and door = 90 + 19.5
= 109.5 ft²
Area of four walls except for windows and door = Area of four walls - Total area of windows and door
= 810 - 109.5
= 700.5 ft²
Cost of coloring per square foot = Rs 300
Total cost of coloring 700.5 ft² = Rs 700.5 × 300
= Rs 210,150
Hence, the total cost for coloring the four walls is Rs 210,150.
6.
The height of a square room is 8 ft. The total cost for laying carpet on that room at the rate of 200 per square ft is 39,200. Find the cost for coloring on the four walls of the room at the rate of 350 per square ft.
Solution:
Here in the square room,
Height of the room (h) = 8 ft
Total cost of laying carpet = Rs 39,200
Cost of carpet per square foot = Rs 200
We know that,
Area of the floor = Total cost / Cost per square foot
= 39,200 / 200
= 196 ft²
Since the room is square, let the side length be a ft,
Area of the floor = a × a
So, a² = 196
or, a = √196
or, a = 14 ft
Therefore, length (l) = 14 ft and breadth (b) = 14 ft
Area of the four walls = 2h(l + b)
= 2 × 8 × (14 + 14)
= 16 × 28
= 448 ft²
Cost of coloring per square foot = Rs 350
Total cost of coloring 448 ft² = Rs 448 × 350
= Rs 156,800
Hence, the cost for coloring the four walls of the room is Rs 156,800.
7.
A school pays Rs 7,056 for laying carpet at the rate of 144 per square meter on a square room for teachers and staff. Except for the windows and doors of the same room, when the paper is pasted on four walls at the rate of 400 per square meter a total of Rs 30,000 is needed. If the total area of windows and doors is 9 m², find the height of the room.
Solution:
Here in the square room,
Total cost of laying carpet = Rs 7,056
Cost of carpet per square meter = Rs 144
We know that,
Area of the floor = Total cost / Cost per square meter
= 7,056 / 144
= 49 m²
Since the room is square, let the side length be a meters,
Area of the floor = a × a
So, a² = 49
or, a = √49
or, a = 7 m
Therefore, length (l) = 7 m and breadth (b) = 7 m
Now, for the walls,
Total cost of pasting paper = Rs 30,000
Cost of pasting paper per square meter = Rs 400
Area of four walls except windows and doors = Total cost / Cost per square meter
= 30,000 / 400
= 75 m²
Total area of windows and doors = 9 m²
Let the height of the room be h meters,
Area of the four walls = 2h(l + b)
= 2h(7 + 7)
= 2h × 14
= 28h m²
Total area of four walls except windows and doors = Area of four walls - Area of windows and doors
So, 28h - 9 = 75
or, 28h = 75 + 9
or, 28h = 84
or, h = 84 / 28
or, h = 3 m
Hence, the height of the room is 3 m.
8.
In Nima’s house, the length of the room is double it’s breadth and height of a room is 2.8 meters. Apart from the window, door, and ceiling of the room, Rs 54,000 is needed to draw the picture on its four walls at the rate of Rs 900 per square meter. If the area of windows and doors is 7.2 square meters, find out the total cost of laying tiles on the floor of the room at the rate of Rs 200 per square meter.
Solution:
Here in the rectangular room,
Height of the room (h) = 2.8 m
Let the breadth of the room be b meters,
Then, length of the room (l) = 2b meters (since length is double the breadth)
Total cost of drawing pictures = Rs 54,000
Cost of drawing per square meter = Rs 900
Area of four walls except windows and doors = Total cost / Cost per square meter
= 54,000 / 900
= 60 m²
Total area of windows and doors = 7.2 m²
Area of the four walls = 2h(l + b)
= 2 × 2.8 × (2b + b)
= 2 × 2.8 × 3b
= 16.8b m²
Area of four walls except windows and doors = Area of four walls - Area of windows and doors
So, 16.8b - 7.2 = 60
or, 16.8b = 60 + 7.2
or, 16.8b = 67.2
or, b = 67.2 / 16.8
or, b = 4 m
Therefore, length (l) = 2b = 2 × 4 = 8 m
Area of the floor = l × b
= 8 × 4
= 32 m²
Cost of laying tiles per square meter = Rs 200
Total cost of laying tiles on 32 m² = Rs 32 × 200
= Rs 6,400
Hence, the total cost of laying tiles on the floor of the room is Rs 6,400.
9.
The length of a school meeting hall is twice the width and the width is twice the height. The school pays Rs 43,200 for coloring its four walls including windows and doors at the rate of 225 per square meter. Find the total cost for laying tiles on the floor of the room at the rate of 250 per square meter.
Solution:
Here in the rectangular meeting hall,
Let the height of the hall be h meters,
Then, width (b) = 2h (since width is twice the height)
And, length (l) = 2b = 2 × 2h = 4h (since length is twice the width)
Total cost of coloring = Rs 43,200
Cost of coloring per square meter = Rs 225
Area of the four walls including windows and doors = Total cost / Cost per square meter
= 43,200 / 225
= 192 m²
Area of the four walls = 2h(l + b)
= 2h(4h + 2h)
= 2h × 6h
= 12h² m²
Since the area includes windows and doors,
12h² = 192
or, h² = 192 / 12
or, h² = 16
or, h = √16
or, h = 4 m
Therefore, height (h) = 4 m
Breadth (b) = 2h = 2 × 4 = 8 m
Length (l) = 4h = 4 × 4 = 16 m
Area of the floor = l × b
= 16 × 8
= 128 m²
Cost of laying tiles per square meter = Rs 250
Total cost of laying tiles on 128 m² = Rs 128 × 250
= Rs 32,000
Hence, the total cost for laying tiles on the floor of the room is Rs 32,000.
10.
The size of a park decorated by a municipality is 50 m × 40 m. If the square slab of dubo sized 20 cm is laid,
(a) Find, how many slabs are needed?
(b) Find the total cost for laying slabs if the cost for laying per slab is Rs. 225.
(c) After a few years, considering the park to be small, the municipality decided to increase the length and width by 15 m and 10 m respectively. Find out how many slabs need to be added to cover the square of the same length in the increased area and how much it costs at the same rate.
Solution:
(a) Find how many slabs are needed?
Area of the park = Length × Width
Here, Length = 50 m, Width = 40 m
So, Area of the park = 50 × 40 = 2000 m²
Area of one square slab = Side × Side
Since the slab is 20 cm × 20 cm, convert cm to meters: 20 cm = 20/100 = 0.2 m
Area of one slab = 0.2 × 0.2 = 0.04 m²
Number of slabs needed = Area of the park / Area of one slab
= 2000 / 0.04
= 50,000 slabs
Hence, 50,000 slabs are needed.
(b) Find the total cost for laying slabs if the cost for laying per slab is Rs. 225.
Number of slabs = 50,000 (from part (a))
Cost of laying per slab = Rs. 225
Total cost = Number of slabs × Cost per slab
= 50,000 × 225
= 50,000 × (200 + 25)
= (50,000 × 200) + (50,000 × 25)
= 10,000,000 + 1,250,000
= 11,250,000
Hence, the total cost for laying slabs is Rs. 11,250,000.
(c) After a few years, considering the park to be small, the municipality decided to increase the length and width by 15 m and 10 m respectively. Find out how many slabs need to be added to cover the square of the same length in the increased area and how much it costs at the same rate.
New length = Original length + 15 m = 50 + 15 = 65 m
New width = Original width + 10 m = 40 + 10 = 50 m
New area of the park = New length × New width
= 65 × 50 = 3250 m²
Original area of the park = 2000 m² (from part (a))
Increased area = New area - Original area
= 3250 - 2000 = 1250 m²
Number of slabs for the increased area = Increased area / Area of one slab
Area of one slab = 0.04 m² (from part (a))
Number of slabs = 1250 / 0.04 = 31,250 slabs
Cost of laying additional slabs = Number of slabs × Cost per slab
= 31,250 × 225
= 31,250 × (200 + 25)
= (31,250 × 200) + (31,250 × 25)
= 6,250,000 + 781,250
= 7,031,250
Hence, 31,250 slabs need to be added, and the cost for these additional slabs is Rs. 7,031,250.
Vedanta Excel in Mathematics - Book 9
Unit 5
Area
EXERCISE 5.1 General Section
1. Fill in the blanks with correct answers.Solution:
a) What is the area of a triangle whose base is x cm and height is y cm?
Area = ½ × base × height = ½ × x × y cm²= ½ xy cm²
b) Write down the area of a right-angled triangle in which base is m cm and perpendicular is n cm.
Area = ½ × base × perpendicular = ½ × m × n cm²
c) What is the area of an equilateral triangle whose side is ‘x’ cm?
Area = (√3/4) × side² = (√3/4) × x² cm²
d) If the length of each equal side of an isosceles triangle is p cm and base is q cm, what is its area?
Area = (q/4) × √(4p² - q²) cm²
e) The sides of a triangle are x, y, and z units respectively, what is the semi-perimeter of the triangle?
Semi-perimeter = (x + y + z)/2 units
f) The semi-perimeter of a triangle having side lengths a cm, b cm, and c cm is s cm. Find its area.
Area = √s(s - a)(s - b)(s - c)cm²
g) What is the area of a triangle whose sides are p cm, q cm, and r cm?
Semi-perimeter (s) = (p + q + r)/2
Area = √s(s - p)(s - q)(s - r) cm²
2. Find the area of the following triangles.2. a) Find the area of the following triangles.
(i) Triangle ABC with sides AB = 8 cm, BC = 6 cm, CA = 10 cm
Solution:
Here, the sides of triangle ABC are a = 8 cm, b = 6 cm, c = 10 cm.
Semi-perimeter (s) = (8 cm + 6 cm + 10 cm) / 2 = 24 cm / 2 = 12 cm
= √{(-1) - 4}² + {5 - (-7)}²
Area = √[s(s - a)(s - b)(s - c)]
= √[12(12 - 8)(12 - 6)(12 - 10)]
= √[12 × 4 × 6 × 2]
= √[576]
= 24 cm²
Hence, the area of triangle ABC is 24 cm².
(ii) Triangle PQR with sides PQ = 13 cm, QR = 14 cm, RP = 15 cm
Solution:
Here, the sides of triangle PQR are a = 13 cm, b = 14 cm, c = 15 cm.
Semi-perimeter (s) = (13 cm + 14 cm + 15 cm) / 2 = 42 cm / 2 = 21 cm
Area = √[s(s - a)(s - b)(s - c)]
= √[21(21 - 13)(21 - 14)(21 - 15)]
= √[21 × 8 × 7 × 6]
= √[7056]
= 84 cm²
Hence, the area of triangle PQR is 84 cm².
(iii) Triangle XYZ with sides XY = 25 cm, YZ = 17 cm, ZX = 12 cm
Solution:
Here, the sides of triangle XYZ are a = 25 cm, b = 17 cm, c = 12 cm.
Semi-perimeter (s) = (25 cm + 17 cm + 12 cm) / 2 = 54 cm / 2 = 27 cm
Area = √[s(s - a)(s - b)(s - c)]
= √[27(27 - 25)(27 - 17)(27 - 12)]
= √[27 × 2 × 10 × 15]
= √[8100]
= 90 cm²
Hence, the area of triangle XYZ is 90 cm².
(iv) Triangle LMN with sides LM = 41 cm, MN = 18 cm, NL = 41 cm
Solution:
Here, the sides of triangle LMN are a = 41 cm, b = 18 cm, c = 41 cm.
Semi-perimeter (s) = (41 cm + 18 cm + 41 cm) / 2 = 100 cm / 2 = 50 cm
Area = √[s(s - a)(s - b)(s - c)]
= √[50(50 - 41)(50 - 18)(50 - 41)]
= √[50 × 9 × 32 × 9]
= √[129600]
= 360 cm²
Hence, the area of triangle LMN is 360 cm².
b) The edges of a triangular kitchen garden are 20 ft, 21 ft, and 29 ft long. Calculate the area of the garden.
Solution:
Here, the sides of the triangular garden are a = 20 ft, b = 21 ft, and c = 29 ft.
Now, semi-perimeter (s) = (20 ft + 21 ft + 29 ft)/2 = 70/2 = 35 ft
Area of the triangle = √[s(s - a)(s - b)(s - c)]
= √[35(35 - 20)(35 - 21)(35 - 29)]
= √[35 × 15 × 14 × 6]
= √[44100]
= 210
Hence, the area of the garden is 210 ft².
c) The sides of the triangular park are 193 m, 194 m, and 195 m. Find its area.
Solution:
Here, the sides of the triangular park are a = 193 m, b = 194 m, and c = 195 m.
Now, semi-perimeter (s) = (193 m + 194 m + 195 m)/2 = 582/2 = 291 m
Area of the triangle = √[s(s - a)(s - b)(s - c)]
= √[291(291 - 193)(291 - 194)(291 - 195)]
= √[291 × 98 × 97 × 96]
= √[265,471,936]
= 16296 m²
Hence, the area of the park is 16296 m².
d) Find the area of an isosceles triangle in which each of the equal sides is 25 cm and the base is 14 cm.
Solution:
Here, in an isosceles triangle, equal sides (a) = 25 cm and base (b) = 14 cm.
Area of isosceles triangle = (b/4) × √[4a² - b²]
= (14/4) × √[4(25)² - 14²]
= 3.5 × √[4 × 625 - 196]
= 3.5 × √[2500 - 196]
= 3.5 × √[2304]
= 3.5 × 48
= 168 cm²
Hence, the area of the isosceles triangle is 168 cm².
e) Find the area of an equilateral triangle whose perimeter is 12 cm.
Solution:
Here, the perimeter of the equilateral triangle = 12 cm
Perimeter = 3a = 12 cm
Side (a) = 12/3 = 4 cm
Area of equilateral triangle = (√3/4) × side²
= (√3/4) × 4²
= (√3/4) × 16
= 4√3 cm²
Hence, the area of the equilateral triangle is 4√3 cm².
Creative Section - A
3. Find the area of the following quadrilaterals.
a) In Quadrilateral ABCD, AB = CD = 13 cm, BC = AD = 14 cm, and diagonal AC = 15 cm
Solution:
Here, quadrilateral ABCD can be divided into two scalene triangles, ΔABC and ΔADC, by the diagonal AC.
In ΔABC, AB = c = 13 cm, BC = a = 14 cm, AC = b = 15 cm
Semi-perimeter (s) = (a + b + c)/2 = (14 + 15 + 13)/2 = 42/2 = 21 cm
Area of ΔABC (A₁) = √[s(s – a)(s – b)(s – c)]
= √[21(21 – 14)(21 – 15)(21 – 13)] cm²
= √[21 × 7 × 6 × 8] cm²
= √[7056] cm²
= 84 cm² (since √7056 = 84)
Again in ΔADC, AD = a = 14 cm, DC = b = 13 cm, AC = c = 15 cm
Semi-perimeter (s) = (a + b + c)/2 = (14 + 13 + 15)/2 = 42/2 = 21 cm
Area of ΔADC (A₂) = √[s(s – a)(s – b)(s – c)]
= √[21(21 – 14)(21 – 13)(21 – 15)] cm²
= √[21 × 7 × 8 × 6] cm²
= √[7056] cm²
= 84 cm² (since √7056 = 84)
Area of quadrilateral ABCD = A₁ + A₂
= (84 + 84) cm²
= 168 cm²
Hence, the area of quadrilateral ABCD is 168 cm².
b) In Quadrilateral PQRS, PQ = QR = RS = SP = 25 cm and diagonal = 48 cm
Solution:
Here, quadrilateral PQRS has all sides equal (PQ = QR = RS = SP = 25 cm), indicating it is a rhombus. The given diagonal (48 cm) is one of the diagonals of the rhombus, say PR. The area of a rhombus can be calculated using the formula involving its diagonals: Area = (1/2) × d₁ × d₂, where d₁ and d₂ are the diagonals. We need to find the length of the other diagonal QS.
In a rhombus, the diagonals bisect each other at right angles. Let PR = 48 cm and QS = d₂. The diagonals split the rhombus into four right-angled triangles. Consider one such triangle (e.g., ΔPOR, where O is the intersection of PR and QS):
PO = PR/2 = 48/2 = 24 cm, OQ = QS/2 = d₂/2, and PQ = 25 cm (side of the rhombus).
Using the Pythagorean theorem in ΔPOR:
PQ² = PO² + OQ²
25² = 24² + (d₂/2)²
625 = 576 + (d₂/2)²
(d₂/2)² = 625 - 576 = 49
d₂/2 = √49 = 7
d₂ = 14 cm
Now, the area of the rhombus PQRS = (1/2) × d₁ × d₂
= (1/2) × 48 cm × 14 cm
= (1/2) × 672 cm²
= 336 cm²
Hence, the area of quadrilateral PQRS is 336 cm².
c) In Quadrilateral ABCD, AB = 9 cm, BC = 10 cm, CD = 25 cm, DA = 12 cm, and diagonal AC = 17 cm
Solution:
Here, quadrilateral ABCD can be divided into two scalene triangles, ΔABC and ΔADC, by the diagonal AC.
In ΔABC, AB = c = 9 cm, BC = a = 10 cm, AC = b = 17 cm
Semi-perimeter (s) = (a + b + c)/2 = (10 + 17 + 9)/2 = 36/2 = 18 cm
Area of ΔABC (A₁) = √[s(s – a)(s – b)(s – c)]
= √[18(18 – 10)(18 – 17)(18 – 9)] cm²
= √[18 × 8 × 1 × 9] cm²
= √[1296] cm²
= 36 cm²
Again in ΔADC, AD = a = 12 cm, DC = b = 25 cm, AC = c = 17 cm
Semi-perimeter (s) = (a + b + c)/2 = (12 + 25 + 17)/2 = 54/2 = 27 cm
Area of ΔADC (A₂) = √[s(s – a)(s – b)(s – c)]
= √[27(27 – 12)(27 – 25)(27 – 17)] cm²
= √[27 × 15 × 2 × 10] cm²
= √[8100] cm²
= 90 cm²
Area of quadrilateral ABCD = A₁ + A₂
= (36 + 90) cm²
= 126 cm²
Hence, the area of quadrilateral ABCD is 126 cm².
d) In Quadrilateral ABCD, AB = AD = 13 cm, BC = 21 cm, CD = 11 cm, and diagonal AC = 20 cm
Solution:
Here, quadrilateral ABCD can be divided into two scalene triangles, ΔABC and ΔADC, by the diagonal AC.
In ΔABC, AB = c = 13 cm, BC = a = 21 cm, AC = b = 20 cm
Semi-perimeter (s) = (a + b + c)/2 = (21 + 20 + 13)/2 = 54/2 = 27 cm
Area of ΔABC (A₁) = √[s(s – a)(s – b)(s – c)]
= √[27(27 – 21)(27 – 20)(27 – 13)] cm²
= √[27 × 6 × 7 × 14] cm²
= √[15876] cm²
= √[81 × 196] cm²
= 9 × 14 cm²
= 126 cm²
Again in ΔADC, AD = a = 13 cm, DC = b = 11 cm, AC = c = 20 cm
Semi-perimeter (s) = (a + b + c)/2 = (13 + 11 + 20)/2 = 44/2 = 22 cm
Area of ΔADC (A₂) = √[s(s – a)(s – b)(s – c)]
= √[22(22 – 13)(22 – 11)(22 – 20)] cm²
= √[22 × 9 × 11 × 2] cm²
= √[4356] cm²
= √[36 × 121] cm²
= 6 × 11 cm²
= 66 cm²
Area of quadrilateral ABCD = A₁ + A₂
= (126 + 66) cm²
= 192 cm²
Hence, the area of quadrilateral ABCD is 192 cm².
e) In Quadrilateral WXYZ, WX = XY = XZ (Diagonal) = 17 cm, XY = 30 cm, WZ = 17 cm, and YZ = 16 cm
Solution:
Here, in ΔWXZ, WX = c = 17 cm, WZ = a = 17 cm, XZ = b = 17 cm
Semi-perimeter (s) = (a + b + c)/2 = (17 + 17 + 17)/2 = 51/2 = 25.5 cm
Area of ΔWXZ (A₁) = √[s(s – a)(s – b)(s – c)]
= √[25.5(25.5 – 17)(25.5 – 17)(25.5 – 17)] cm²
= √[25.5 × 8.5 × 8.5 × 8.5] cm²
= √[25.5 × (8.5)³] cm²
= √[25.5 × 614.125] cm²
= √[15660.1875] cm²
≈ 125.14 cm²
In ΔXYZ, XY = a = 30 cm, YZ = b = 16 cm, XZ = c = 17 cm
Semi-perimeter (s) = (a + b + c)/2 = (30 + 16 + 17)/2 = 63/2 = 31.5 cm
Area of ΔXYZ (A₂) = √[s(s – a)(s – b)(s – c)]
= √[31.5(31.5 – 30)(31.5 – 16)(31.5 – 17)] cm²
= √[31.5 × 1.5 × 15.5 × 14.5] cm²
= √[10631.625] cm²
≈ 103.11 cm² (since √10631.625 ≈ 103.11)
Area of quadrilateral WXYZ = A₁ + A₂
= (125.14 + 103.11) cm²
= 228.25 cm²
Hence, the area of quadrilateral WXYZ is 228.25 cm².
f) In Quadrilateral ABCD, AB = 16 cm, BC = 13 cm, CD = 21 cm, and AD = 12 cm
4. Solve the following problems.a) If the ratio of the sides of a triangle is 3:4:5 and its perimeter is 36 cm, find: (i) the measure of its sides (ii) its area
Solution:
Here, the perimeter of the triangle = 36 cm
Let the sides of the triangle be 3x cm, 4x cm, and 5x cm.
(i) Now, 3x + 4x + 5x = 36 cm
12x = 36 cm
or, x = 36/12 = 3 cm
Sides = 3x = 3 × 3 = 9 cm, 4x = 4 × 3 = 12 cm, 5x = 5 × 3 = 15 cm
Hence, the sides are 9 cm, 12 cm, and 15 cm.
(ii) Semi-perimeter (s) = 36/2 = 18 cm
Area of the triangle = √[s(s - a)(s - b)(s - c)]
= √[18(18 - 9)(18 - 12)(18 - 15)]
= √[18 × 9 × 6 × 3]
= √[2916]
= 54 cm²
Hence, the area of the triangle is 54 cm².
b) The perimeter of a triangular park is 84 m and the ratio of its sides is 13:14:15, find: (i) the measure of its sides (ii) its area
Solution:
Here, the perimeter of the triangular park = 84 m
Let the sides of the triangle be 13x m, 14x m, and 15x m.
(i) Now, 13x + 14x + 15x = 84 m
42x = 84 m
x = 84/42 = 2 m
Sides = 13x = 13 × 2 = 26 m, 14x = 14 × 2 = 28 m, 15x = 15 × 2 = 30 m
Hence, the sides are 26 m, 28 m, and 30 m.
(ii) Semi-perimeter (s) = 84/2 = 42 m
Area of the triangle = √[s(s - a)(s - b)(s - c)]
= √[42(42 - 26)(42 - 28)(42 - 30)]
= √[42 × 16 × 14 × 12]
= √[112,896]
= 336 m²
Hence, the area of the park is 336 m².
c) The ratio of the sides of a triangular field is 12:5:13 and its area is 120 m², find the perimeter of the field.
Solution:
Let the sides of the triangular field be 12x m, 5x m, and 13x m.
Now, semi-perimeter (s) = (12x + 5x + 13x)/2 = 30x/2 = 15x m
Area of the triangle = √[s(s - a)(s - b)(s - c)]
or, 120 = √[15x(15x - 12x)(15x - 5x)(15x - 13x)]
or, 120 = √[15x × 3x × 10x × 2x]
or, 120 = √[900x⁴]
or, 120 = 30x²
or, x² = 120/30 = 4
or, x = √4 = 2
Sides = 12x = 12 × 2 = 24 m, 5x = 5 × 2 = 10 m, 13x = 13 × 2 = 26 m
Perimeter = 24 + 10 + 26 = 60 m
Hence, the perimeter of the field is 60 m.
d) The area of a triangular kitchen garden is 324 sq. ft and its edges are in the ratio 9:10:17. Find the perimeter of the garden.
Solution:
Let the sides of the triangular garden be 9x ft, 10x ft, and 17x ft.
Now, semi-perimeter (s) = (9x + 10x + 17x)/2 = 36x/2 = 18x ft
Area of the triangle = √[s(s - a)(s - b)(s - c)]
or, 324 = √[18x(18x - 9x)(18x - 10x)(18x - 17x)]
or, 324 = √[18x × 9x × 8x × x]
or, 324 = √[1296x⁴]
or, 324 = 36x²
or, x² = 324/36 = 9
or, x = √9 = 3
Sides = 9x = 9 × 3 = 27 ft, 10x = 10 × 3 = 30 ft, 17x = 17 × 3 = 51 ft
Perimeter = 27 + 30 + 51 = 108 ft
Hence, the perimeter of the garden is 108 ft.
Creative Section - B
5. Solve the following problems.
a) The shape of a piece of land is a parallelogram whose adjacent sides are 12 m and 17 m and the corresponding diagonal is 25 m. Find the area of the land.
Solution:
Here, the parallelogram has adjacent sides AB = 12 m and AD = 17 m, and the corresponding diagonal BD = 25 m.
In ΔABD, the sides are AB = 12 m, AD = 17 m, and BD = 25 m.
Now, semi-perimeter (s) = (12 m + 17 m + 25 m)/2 = 54/2 = 27 m
Area of ΔABD = √[s(s - a)(s - b)(s - c)]
= √[27(27 - 12)(27 - 17)(27 - 25)]
= √[27 × 15 × 10 × 2]
= √[8100]
= 90 m²
Area of parallelogram = 2 × Area of ΔABD = 2 × 90 m² = 180 m²
Hence, the area of the land is 180 m².
b) A park is in the shape of a parallelogram. The lengths of two adjacent edges are 51 m and 52 m and the corresponding diagonal is 53 m, what is the area of the park?
Solution:
Here, the parallelogram has adjacent sides AB = 51 m and AD = 52 m, and the corresponding diagonal BD = 53 m.
In ΔABD, the sides are AB = 51 m, AD = 52 m, and BD = 53 m.
Now, semi-perimeter (s) = (51 m + 52 m + 53 m)/2 = 156/2 = 78 m
Area of ΔABD = √[s(s - a)(s - b)(s - c)]
= √[78(78 - 51)(78 - 52)(78 - 53)]
= √[78 × 27 × 26 × 25]
= √[1,368,900]
= 1,170 m²
Area of parallelogram = 2 × Area of ΔABD = 2 × 1170 m² = 2340 m²
Hence, the area of the park is 2340 m².
c) A garden is in the shape of a rhombus whose each side is 15 m and one of the two diagonals is 24 m. Find the area of the garden.
Solution:
Here, the rhombus has each side = 15 m and one diagonal (say AC) = 24 m.
In a rhombus, diagonals bisect each other at right angles. Let the other diagonal be BD = 2x m.
In ΔABC, AB = 15 m (side), AC = 24 m (diagonal), and BC = 15 m (side).
Half of diagonal AC = AO = 24/2 = 12 m.
In right-angled ΔAOB, by Pythagoras theorem:
AB² = AO² + BO²
15² = 12² + BO²
225 = 144 + BO²
BO² = 225 - 144 = 81
BO = √81 = 9 m
Full diagonal BD = 2 × BO = 2 × 9 = 18 m
Area of rhombus = (1/2) × diagonal1 × diagonal2
= (1/2) × 24 × 18
= 216 m²
Hence, the area of the garden is 216 m².
d) A farmer has a field in the shape of a rhombus. The perimeter of the field is 400 m and one of its diagonals is 120 m. Find the area of the field in hectares. (1 hectare = 10,000 m²)
Solution:
Here, the perimeter of the rhombus = 400 m and one diagonal (say AC) = 120 m.
Perimeter = 4 × side = 400 m
Side = 400/4 = 100 m
In a rhombus, diagonals bisect each other at right angles. Let the other diagonal be BD = 2x m.
In ΔABC, AB = 100 m (side), AC = 120 m (diagonal), and BC = 100 m (side).
Half of diagonal AC = AO = 120/2 = 60 m.
In right-angled ΔAOB, by Pythagoras theorem:
AB² = AO² + BO²
100² = 60² + BO²
10000 = 3600 + BO²
BO² = 10000 - 3600 = 6400
BO = √6400 = 80 m
Full diagonal BD = 2 × BO = 2 × 80 = 160 m
Area of rhombus = (1/2) × diagonal1 × diagonal2
= (1/2) × 120 × 160
= 9600 m²
Area in hectares = 9600/10000 = 0.96 hectares
Hence, the area of the field is 0.96 hectares.
6. Solve the following problems.
a) An umbrella is made by stitching 8 triangular pieces of cloth of different colors as shown in the figure. Each piece measures 50 cm, 50 cm, and 28 cm.
(i) How much cloth of each color is required for the umbrella?
(ii) If the rate of cost of the cloth is Rs 0.15 per sq. cm, find the cost of the cloths required to make the umbrella.
Solution:
Here, each triangular piece has sides a = 50 cm, b = 50 cm, and c = 28 cm.
(i) Now, semi-perimeter (s) = (50 cm + 50 cm + 28 cm)/2 = 128/2 = 64 cm
Area of one triangular piece = √[s(s - a)(s - b)(s - c)]
= √[64(64 - 50)(64 - 50)(64 - 28)]
= √[64 × 14 × 14 × 36]
= √[451,584]
= 672 cm²
Area of cloth for 8 pieces = 8 × 672 cm² = 5376 cm²
Since there are 8 different colors, cloth of each color = 5376/8 = 672 cm²
Hence, 672 cm² of cloth of each color is required for the umbrella.
(ii) Total area of cloth = 5376 cm²
Cost of cloth = Area × Rate = 5376 × Rs 0.15 = Rs 806.40
Hence, the cost of the cloths required to make the umbrella is Rs 806.40.
b) Mrs. Limbu made a rangoli which has 8 identical triangles. Each triangle measures 20 cm, 13 cm, and 11 cm.
(i) How much surface of the floor is covered with the design?
(ii) If the rate of cost of the color is Rs 2.50 per sq. cm, find the cost of the colors required to make the design.
Solution:
Here, each triangular piece has sides a = 20 cm, b = 13 cm, and c = 11 cm.
(i) Now, semi-perimeter (s) = (20 cm + 13 cm + 11 cm)/2 = 44/2 = 22 cm
Area of one triangular piece = √[s(s - a)(s - b)(s - c)]
= √[22(22 - 20)(22 - 13)(22 - 11)]
= √[22 × 2 × 9 × 11]
= √[4356]
= 66 cm²
Total area of 8 triangles = 8 × 66 cm² = 528 cm²
Hence, the surface of the floor covered with the design is 528 cm².
(ii) Cost of color = Area × Rate = 528 × Rs 2.50 = Rs 1320
Hence, the cost of the colors required to make the design is Rs 1320.
c) A floral design on a floor of a room is made up of 16 triangular pieces of tiles. The sides of each piece of tile are 28 cm, 15 cm, and 41 cm.
(i) Find the area of the design.
(ii) Estimate the cost of polishing the tiles at Rs 1.50 per sq. cm.
Solution:
Here, each triangular piece has sides a = 28 cm, b = 15 cm, and c = 41 cm.
(i) Now, semi-perimeter (s) = (28 cm + 15 cm + 41 cm)/2 = 84/2 = 42 cm
Area of one triangular piece = √[s(s - a)(s - b)(s - c)]
= √[42(42 - 28)(42 - 15)(42 - 41)]
= √[42 × 14 × 27 × 1]
= √[15,876]
= 126 cm²
Total area of 16 triangles = 16 × 126 cm² = 2016 cm²
Hence, the area of the design is 2016 cm².
(ii) Cost of polishing = Area × Rate = 2016 × Rs 1.50 = Rs 3024
Hence, the cost of polishing the tiles is Rs 3024.
Local land measurement Units in Nepal
EXERCISE 5.2 General Section
1. Answer the following questions.
a) How many Aana are there in 1 Ropani?
Solution:
1 Ropani = 16 Aana
Hence, there are 16 Aana in 1 Ropani.
b) How many Paisa are there in (i) 1 Aana? (ii) 1 Ropani?
Solution:
(i) 1 Aana = 4 Paisa
Hence, there are 4 Paisa in 1 Aana.
(ii) 1 Ropani = 16 Aana
= 16 × 4 Paisa [∵ 1 Aana = 4 Paisa]
= 64 Paisa
Hence, there are 64 Paisa in 1 Ropani.
c) How many Daam are there in (i) 1 Paisa? (ii) 1 Aana? (iii) 1 Ropani?
Solution:
(i) 1 Paisa = 4 Daam
Hence, there are 4 Daam in 1 Paisa.
(ii) 1 Aana = 4 Paisa
= 4 × 4 Daam [∵ 1 Paisa = 4 Daam]
= 16 Daam
Hence, there are 16 Daam in 1 Aana.
(iii) 1 Ropani = 16 Aana
= 16 × 16 Daam [∵ 1 Aana = 16 Daam]
= 256 Daam
Hence, there are 256 Daam in 1 Ropani.
d) How many Kattha are there in 1 Bigha?
Solution:
1 Bigha = 20 Kattha
Hence, there are 20 Kattha in 1 Bigha.
e) How many Dhur are there in (i) 1 Kattha? (ii) 1 Bigha?
Solution:
(i) 1 Kattha = 20 Dhur
Hence, there are 20 Dhur in 1 Kattha.
(ii) 1 Bigha = 20 Kattha
= 20 × 20 Dhur [∵ 1 Kattha = 20 Dhur]
= 400 Dhur
Hence, there are 400 Dhur in 1 Bigha.
f) How many Ropani are there in 1 Bigha?
Solution:
1 Bigha = 13.31 Ropani
Hence, there are 13.31 Ropani in 1 Bigha.
2. Convert the following area of the lands into the units as indicated.a)
(i) 6 Ropani into Aana
Solution:
6 Ropani = 6 × 16 Aana [∵ 1 Ropani = 16 Aana]
= 96 Aana
Hence, 6 Ropani is equal to 96 Aana.
(ii) 10 Ropani 7 Aana into Aana
Solution:
10 Ropani 7 Aana = 10 × 16 Aana + 7 Aana [∵ 1 Ropani = 16 Aana]
= 160 Aana + 7 Aana
= 167 Aana
Hence, 10 Ropani 7 Aana is equal to 167 Aana.
(iii) 12 Ropani 10 Aana 2 Paisa into Paisa
Solution:
12 Ropani 10 Aana 2 Paisa = 12 × 16 Aana + 10 Aana + 2 Paisa [∵ 1 Ropani = 16 Aana]
= 192 Aana + 10 Aana + 2 Paisa
= 202 Aana + 2 Paisa
= 202 × 4 Paisa + 2 Paisa [∵ 1 Aana = 4 Paisa]
= 808 Paisa + 2 Paisa
= 810 Paisa
Hence, 12 Ropani 10 Aana 2 Paisa is equal to 810 Paisa.
(iv) 20 Ropani 15 Aana 3 Paisa into Paisa
Solution:
20 Ropani 15 Aana 3 Paisa = 20 × 16 Aana + 15 Aana + 3 Paisa [∵ 1 Ropani = 16 Aana]
= 320 Aana + 15 Aana + 3 Paisa
= 335 Aana + 3 Paisa
= 335 × 4 Paisa + 3 Paisa [∵ 1 Aana = 4 Paisa]
= 1340 Paisa + 3 Paisa
= 1343 Paisa
Hence, 20 Ropani 15 Aana 3 Paisa is equal to 1343 Paisa.
(v) 5 Bigha into Kattha
Solution:
5 Bigha = 5 × 20 Kattha [∵ 1 Bigha = 20 Kattha]
= 100 Kattha
Hence, 5 Bigha is equal to 100 Kattha.
(vi) 6 Bigha 9 Kattha into Kattha
Solution:
6 Bigha 9 Kattha = 6 × 20 Kattha + 9 Kattha [∵ 1 Bigha = 20 Kattha]
= 120 Kattha + 9 Kattha
= 129 Kattha
Hence, 6 Bigha 9 Kattha is equal to 129 Kattha.
(vii) 7 Bigha 7 Kattha 7 Dhur into Dhur
Solution:
7 Bigha 7 Kattha 7 Dhur = 7 × 20 Kattha + 7 Kattha + 7 Dhur [∵ 1 Bigha = 20 Kattha]
= 140 Kattha + 7 Kattha + 7 Dhur
= 147 Kattha + 7 Dhur
= 147 × 20 Dhur + 7 Dhur [∵ 1 Kattha = 20 Dhur]
= 2940 Dhur + 7 Dhur
= 2947 Dhur
Hence, 7 Bigha 7 Kattha 7 Dhur is equal to 2947 Dhur.
(viii) 8 Bigha 9 Kattha 10 Dhur into Dhur
Solution:
8 Bigha 9 Kattha 10 Dhur = 8 × 20 Kattha + 9 Kattha + 10 Dhur [∵ 1 Bigha = 20 Kattha]
= 160 Kattha + 9 Kattha + 10 Dhur
= 169 Kattha + 10 Dhur
= 169 × 20 Dhur + 10 Dhur [∵ 1 Kattha = 20 Dhur]
= 3380 Dhur + 10 Dhur
= 3390 Dhur
Hence, 8 Bigha 9 Kattha 10 Dhur is equal to 3390 Dhur.
b)
(i) 32 Aana into Ropani
Solution:
32 Aana = 32/16 Ropani [∵ 16 Aana = 1 Ropani]
= 2 Ropani
Hence, 32 Aana is equal to 2 Ropani.
(ii) 3 Ropani 8 Aana into Ropani
Solution:
3 Ropani 8 Aana = 3 Ropani + 8 Aana
= 3 Ropani + 8/16 Ropani [∵ 16 Aana = 1 Ropani]
= 3 Ropani + 0.5 Ropani
= 3.5 Ropani
Hence, 3 Ropani 8 Aana is equal to 3.5 Ropani.
(iii) 2 Ropani 8 Aana 4 Paisa into Ropani
Solution:
2 Ropani 8 Aana 4 Paisa = 2 Ropani + 8 Aana + 4/4 Aana [∵ 4 Paisa = 1 Aana]
= 2 Ropani + 8 Aana + 1 Aana
= 2 Ropani + 9 Aana
= 2 Ropani + 9/16 Ropani [∵ 16 Aana = 1 Ropani]
= 2 Ropani + 0.5625 Ropani
= 2.5625 Ropani
Hence, 2 Ropani 8 Aana 4 Paisa is equal to 2.5625 Ropani.
(iv) 20 Ropani 4 Aana 8 Paisa into Ropani
Solution:
20 Ropani 4 Aana 8 Paisa = 20 Ropani + 4 Aana + 8/4 Aana [∵ 4 Paisa = 1 Aana]
= 20 Ropani + 4 Aana + 2 Aana
= 20 Ropani + 6 Aana
= 20 Ropani + 6/16 Ropani [∵ 16 Aana = 1 Ropani]
= 20 Ropani + 0.375 Ropani
= 20.375 Ropani
Hence, 20 Ropani 4 Aana 8 Paisa is equal to 20.375 Ropani.
(v) 100 Kattha into Bigha
Solution:
100 Kattha = 100/20 Bigha [∵ 20 Kattha = 1 Bigha]
= 5 Bigha
Hence, 100 Kattha is equal to 5 Bigha.
(vi) 2 Bigha 10 Kattha into Kattha
Solution:
2 Bigha 10 Kattha = 2 × 20 Kattha + 10 Kattha [∵ 1 Bigha = 20 Kattha]
= 40 Kattha + 10 Kattha
= 50 Kattha
Hence, 2 Bigha 10 Kattha is equal to 50 Kattha.
(vii) 20 Bigha 10 Kattha 5 Dhur into Bigha
Solution:
20 Bigha 10 Kattha 5 Dhur = 20 Bigha + 10 Kattha + 5/20 Kattha [∵ 20 Dhur = 1 Kattha]
= 20 Bigha + 10 Kattha + 0.25 Kattha
= 20 Bigha + 10.25 Kattha
= 20 Bigha + 10.25/20 Bigha [∵ 20 Kattha = 1 Bigha]
= 20 Bigha + 0.5125 Bigha
= 20.5125 Bigha
Hence, 20 Bigha 10 Kattha 5 Dhur is equal to 20.5125 Bigha.
(viii) 11 Bigha 4 Kattha 10 Dhur into Bigha
Solution:
11 Bigha 4 Kattha 10 Dhur = 11 Bigha + 4 Kattha + 10/20 Kattha [∵ 20 Dhur = 1 Kattha]
= 11 Bigha + 4 Kattha + 0.5 Kattha
= 11 Bigha + 4.5 Kattha
= 11 Bigha + 4.5/20 Bigha [∵ 20 Kattha = 1 Bigha]
= 11 Bigha + 0.225 Bigha
= 11.225 Bigha
Hence, 11 Bigha 4 Kattha 10 Dhur is equal to 11.225 Bigha.
Creative Section
3. Solve the following problems.a) Mr. Majhi has 5 Bigha 4 Kattha land in Jhapa district. Change it into Ropani-Aana-Paisa-Daam format. (1 Bigha = 13.31 Ropani)
Solution:
Here, 5 Bigha 4 Kattha = 5 Bigha + 4 Kattha
= 5 Bigha + 4/20 Bigha [∵ 20 Kattha = 1 Bigha]
= 5 Bigha + 0.2 Bigha
= 5.2 Bigha
Now, convert into Ropani: 5.2 Bigha = 5.2 × 13.31 Ropani [∵ 1 Bigha = 13.31 Ropani]
= 69.212 Ropani
Ropani part = 69 Ropani
Aana part = 0.212 Ropani = 0.212 × 16 Aana [∵ 1 Ropani = 16 Aana]
= 3.392 Aana
Aana part = 3 Aana
Paisa part = 0.392 Aana = 0.392 × 4 Paisa [∵ 1 Aana = 4 Paisa]
= 1.568 Paisa
Paisa part = 1 Paisa
Daam part = 0.568 Paisa = 0.568 × 4 Daam [∵ 1 Paisa = 4 Daam]
= 2.272 Daam
Daam part = 2 Daam (approx.)
Hence, 5 Bigha 4 Kattha is approximately 69 Ropani 3 Aana 1 Paisa 2 Daam.
b) Mr. Jha has 10 Bigha 10 Kattha 10 Dhur plot of land in Dhanusha district. Change it into Ropani-Aana-Paisa-Daam format.
Solution:
Here, 10 Bigha 10 Kattha 10 Dhur = 10 Bigha + 10 Kattha + 10/20 Kattha [∵ 20 Dhur = 1 Kattha]
= 10 Bigha + 10 Kattha + 0.5 Kattha
= 10 Bigha + 10.5 Kattha
= 10 Bigha + 10.5/20 Bigha [∵ 20 Kattha = 1 Bigha]
= 10 Bigha + 0.525 Bigha
= 10.525 Bigha
Now, convert into Ropani: 10.525 Bigha = 10.525 × 13.31 Ropani [∵ 1 Bigha = 13.31 Ropani]
= 140.08775 Ropani
Ropani part = 140 Ropani
Aana part = 0.08775 Ropani = 0.08775 × 16 Aana [∵ 1 Ropani = 16 Aana]
= 1.404 Aana
Aana part = 1 Aana
Paisa part = 0.404 Aana = 0.404 × 4 Paisa [∵ 1 Aana = 4 Paisa]
= 1.616 Paisa
Paisa part = 1 Paisa
Daam part = 0.616 Paisa = 0.616 × 4 Daam [∵ 1 Paisa = 4 Daam]
= 2.464 Daam
Daam part = 2 Daam (approx.)
Hence, 10 Bigha 10 Kattha 10 Dhur is approximately 140 Ropani 1 Aana 1 Paisa 2 Daam.
c) Mr. Deuba has 73 Ropani 8 Aana 2 Paisa 1.664 Daam land in Sudurpaschim province. Change it into Bigha-Kattha-Dhur system.
Solution:
Here, 73 Ropani 8 Aana 2 Paisa 1.664 Daam = 73 Ropani + 8 Aana + 2/4 Aana + 1.664/16 Aana [∵ 4 Paisa = 1 Aana, 4 Daam = 1 Paisa]
= 73 Ropani + 8 Aana + 0.5 Aana + 0.104 Aana [∵ 1.664 Daam = 1.664/4 = 0.416 Paisa = 0.416/4 = 0.104 Aana]
= 73 Ropani + 8.604 Aana
= 73 Ropani + 8.604/16 Ropani [∵ 16 Aana = 1 Ropani]
= 73 Ropani + 0.53775 Ropani
= 73.53775 Ropani
Now, convert into Bigha: 73.53775 Ropani = 73.53775 / 13.31 Bigha [∵ 1 Bigha = 13.31 Ropani]
= 5.5247 Bigha
Bigha part = 5 Bigha
Kattha part = 0.5247 Bigha = 0.5247 × 20 Kattha [∵ 1 Bigha = 20 Kattha]
= 10.494 Kattha
Kattha part = 10 Kattha
Dhur part = 0.494 Kattha = 0.494 × 20 Dhur [∵ 1 Kattha = 20 Dhur]
= 9.88 Dhur
Dhur part = 9 Dhur (approx.)
Hence, 73 Ropani 8 Aana 2 Paisa 1.664 Daam is approximately 5 Bigha 10 Kattha 9 Dhur.
d) Mrs. Gurung has 29 Ropani 13 Aana 0.4864 Daam land in Gandaki province. Change it into Bigha-Kattha-Dhur system.
Solution:
Here, 29 Ropani 13 Aana 0.4864 Daam = 29 Ropani + 13 Aana + 0.4864/16 Aana [∵ 4 Daam = 1 Paisa, 0.4864 Daam = 0.4864/4 = 0.1216 Paisa = 0.1216/4 = 0.0304 Aana]
= 29 Ropani + 13 Aana + 0.0304 Aana
= 29 Ropani + 13.0304 Aana
= 29 Ropani + 13.0304/16 Ropani [∵ 16 Aana = 1 Ropani]
= 29 Ropani + 0.8144 Ropani
= 29.8144 Ropani
Now, convert into Bigha: 29.8144 Ropani = 29.8144 / 13.31 Bigha [∵ 1 Bigha = 13.31 Ropani]
= 2.2392 Bigha
Bigha part = 2 Bigha
Kattha part = 0.2392 Bigha = 0.2392 × 20 Kattha [∵ 1 Bigha = 20 Kattha]
= 4.784 Kattha
Kattha part = 4 Kattha
Dhur part = 0.784 Kattha = 0.784 × 20 Dhur [∵ 1 Kattha = 20 Dhur]
= 15.68 Dhur
Dhur part = 15 Dhur (approx.)
Hence, 29 Ropani 13 Aana 0.4864 Daam is approximately 2 Bigha 4 Kattha 15 Dhur.
4. Study the following relations and solve the questions given below.Unit In square meter In square feet
1 Ropani = 16 Aana = 64 Paisa = 508.72 m² = 5476 sq. ft
1 Bigha = 20 Kattha = 400 Dhur = 6772.63 m² = 72900 sq. ft
a) A rectangular field is 50 m long and 40 m wide. Convert the area of the field into Ropani.
Solution:
Area of rectangular field = length × width = 50 m × 40 m = 2000 m²
1 Ropani = 508.72 m²
Area in Ropani = 2000 / 508.72 = 3.930 Ropani
Hence, the area of the field is approximately 3.930 Ropani.
b) A rectangular field is 185 ft long and 74 ft wide. Convert the area of the field into Aana.
Solution:
Area of rectangular field = length × width = 185 ft × 74 ft = 13690 sq. ft
1 Aana = 342.25 sq. ft [∵ 1 Ropani = 5476 sq. ft, 1 Ropani = 16 Aana, so 5476/16 = 342.25 sq. ft]
Area in Aana = 13690 / 342.25 = 40 Aana
Hence, the area of the field is 40 Aana.
c) The side of a squared ground is 180 m, find its area in Bigha.
Solution:
Area of square ground = side × side = 180 m × 180 m = 32400 m²
1 Bigha = 6772.63 m²
Area in Bigha = 32400 / 6772.63 = 4.784 Bigha
Hence, the area of the ground is approximately 4.784 Bigha.
d) The edges of a triangular kitchen garden are 13 m, 14 m, and 15 m respectively, find the area of the garden in Dhur.
Solution:
Here, the sides of the triangular garden are a = 13 m, b = 14 m, and c = 15 m.
Semi-perimeter (s) = (13 + 14 + 15)/2 = 42/2 = 21 m
Area = √[s(s - a)(s - b)(s - c)]
= √[21(21 - 13)(21 - 14)(21 - 15)]
= √[21 × 8 × 7 × 6]
= √[7056]
= 84 m²
1 Dhur = 16.93 m²
Area in Dhur = 84 / 16.93 = 4.962 Dhur
Hence, the area of the garden is approximately 4.962 Dhur.
5. The map of a few plots of land is shown in the given figure. Measure the dimensions of the plots with a ruler and answer the following questions. 6. Solve the following problems.a) A municipality is building a hospital in the land of the shape of a rhombus having its diagonals 250 m and 480 m respectively in a mountainous district of Nepal.
(i) What is the formula to find the area of a rhombus?
Solution:
Area of a rhombus = (1/2) × diagonal1 × diagonal2
(ii) Find the area of the land covered by the hospital.
Solution:
Area = (1/2) × 250 m × 480 m
= 60000 m²
Hence, the area of the land is 60000 m².
(iii) Convert the area of the land covered by the hospital into Ropani-Aana-Paisa-Daam format.
Solution:
1 Ropani = 508.72 m²
Area in Ropani = 60000 / 508.72 = 117.918 Ropani
Ropani part = 117 Ropani
Aana part = 0.918 Ropani = 0.918 × 16 Aana = 14.688 Aana
Aana part = 14 Aana
Paisa part = 0.688 Aana = 0.688 × 4 Paisa = 2.752 Paisa
Paisa part = 2 Paisa
Daam part = 0.752 Paisa = 0.752 × 4 Daam = 3.008 Daam
Daam part = 3 Daam (approx.)
Hence, the area is approximately 117 Ropani 14 Aana 2 Paisa 3 Daam.
b) Government is building an international airport in the land of the shape of a quadrilateral in Terai region. The longer diagonal of the land is 800 m. The perpendiculars drawn to the longer diagonal from the opposite vertices are 150 m and 200 m.
(i) Write the formula to find the area of a quadrilateral when the perpendiculars with lengths ‘a’ and ‘b’ are drawn from two opposite vertices on the diagonal ‘d’ of a quadrilateral.
Solution:
Area = (1/2) × d × (a + b)
(ii) Find the area of the land covered by the airport.
Solution:
Area = (1/2) × 800 m × (150 m + 200 m)
= (1/2) × 800 × 350
= 140000 m²
Hence, the area of the land is 140000 m².
(iii) Convert the area of the land covered by the airport into Bigha-Kattha-Dhur format.
Solution:
1 Bigha = 6772.63 m²
Area in Bigha = 140000 / 6772.63 = 20.668 Bigha
Bigha part = 20 Bigha
Kattha part = 0.668 Bigha = 0.668 × 20 Kattha = 13.36 Kattha
Kattha part = 13 Kattha
Dhur part = 0.36 Kattha = 0.36 × 20 Dhur = 7.2 Dhur
Dhur part = 7 Dhur (approx.)
Hence, the area is approximately 20 Bigha 13 Kattha 7 Dhur.
c) Mrs. Gosain has a piece of land in the shape of a rhombus in Bhaktapur. The diagonals of the field are 16 m and 15.9 m respectively. He sold the land to Mrs. Ojha at the rate of Rs 25,00,000 per Aana. How much did Mrs. Ojha pay to Mr. Gosain?
Solution:
Area of rhombus = (1/2) × diagonal1 × diagonal2
= (1/2) × 16 m × 15.9 m
= 127.2 m²
1 Aana = 31.80 m²
Area in Aana = 127.2 / 31.80 = 4 Aana
Cost = Area in Aana × Rate = 4 × Rs 25,00,000 = Rs 1,00,00,000
Hence, Mrs. Ojha paid Rs 1,00,00,000 to Mr. Gosain.
Area of 4 walls, floor and ceiling
EXERCISE 5.3
General Section
1. Answer the following questions.a) What is the relation between the areas of floor and ceiling in a rectangular room?
Solution:
In a rectangular room, the floor and ceiling are congruent.
Area of floor = l × b
Area of ceiling = l × b
Hence, the area of the floor is equal to the area of the ceiling.
b) Write the relation between the areas of opposite walls of a rectangular room.
Solution:
In a rectangular room, the opposite walls along the length are congruent, and the opposite walls along the breadth are congruent.
Area of opposite walls along the length = 2lh
Area of opposite walls along the breadth = 2bh
Hence, the area of opposite walls along the length is 2lh, and the area of opposite walls along the breadth is 2bh.
2. Solve the following problems.a) From the adjoining room, write down the area of its following parts.
[Note: No figure is provided, so I assume a rectangular room with length l, breadth b, and height h is implied. I’ll provide formulas based on this.]
(i) Floor
Solution:
Area of floor = l × b
(ii) Ceiling
Solution:
Area of ceiling = l × b
(iii) Floor and ceiling
Solution:
Area of floor and ceiling = l × b + l × b = 2lb
(iv) Opposite walls along the length
Solution:
Area of opposite walls along the length = 2lh
(v) Opposite walls along the breadth
Solution:
Area of opposite walls along the breadth = 2bh
(vi) Four walls
Solution:
Area of four walls = 2h(l + b)
(vii) Walls and floor
Solution:
Area of walls and floor = 2h(l + b) + l × b
(viii) Walls and ceiling
Solution:
Area of walls and ceiling = 2h(l + b) + l × b
b) The perimeter of the floor of a rectangular room is P m and the height of the room is Q m, what is the area of its four walls?
Solution:
Perimeter of the floor (P) = 2(l + b)
Area of four walls = 2h(l + b)
= P × h
Given h = Q m, Area of four walls = P × Q m²
Hence, the area of its four walls is P × Q m².
c) The length of a square room is x m and its height is y m, what is the area of its four walls?
Solution:
For a square room, length = breadth = x m, height = y m
Area of four walls = 2h(l + b)
= 2y(x + x)
= 2y(2x)
= 4xy m²
Hence, the area of its four walls is 4xy m².
d) If the length of a cubical room is a ft., write down the area of its four walls and ceiling.
Solution:
For a cubical room, length = breadth = height = a ft
Area of four walls = 2h(l + b)
= 2a(a + a)
= 2a(2a)
= 4a² ft²
Area of ceiling = l × b = a × a = a² ft²
Area of four walls and ceiling = 4a² + a² = 5a² ft²
Hence, the area of its four walls and ceiling is 5a² ft².
e) In a rectangular room, if the area of the floor is A m² and the area of its four walls is B m², what is the total surface area of the room?
Solution:
Total surface area of the room = Area of floor + Area of four walls + Area of ceiling
Area of floor = A m²
Area of ceiling = A m² [∵ floor and ceiling are congruent]
Area of four walls = B m²
Total surface area = A + B + A = 2A + B m²
Hence, the total surface area of the room is 2A + B m².
3. Solve the following problems.a) The length of a square room is 10 m, find the area of its floor.
Solution:
For a square room, length = breadth = 10 m
Area of floor = l × b
= 10 m × 10 m
= 100 m²
Hence, the area of its floor is 100 m².
b) A square room is 25 ft. long. Find the area of its ceiling.
Solution:
For a square room, length = breadth = 25 ft
Area of ceiling = l × b
= 25 ft × 25 ft
= 625 ft²
Hence, the area of its ceiling is 625 ft².
c) A cubical room is 12 m long. What is the area of its four walls?
Solution:
For a cubical room, length = breadth = height = 12 m
Area of four walls = 2h(l + b)
= 2 × 12 m (12 m + 12 m)
= 24 m × 24 m
= 576 m²
Hence, the area of its four walls is 576 m².
d) If the perimeter of a rectangular room having height 4 m is 36 m, calculate the area of its walls.
Solution:
Perimeter of the floor (P) = 2(l + b) = 36 m
Height (h) = 4 m
Area of four walls = P × h
= 36 m × 4 m
= 144 m²
Hence, the area of its walls is 144 m².
e) If 5 pieces of carpets having dimensions 10 ft. × 6 ft. of each piece are required for carpeting the floor of a rectangular room, find the area of the floor.
Solution:
Area of one piece of carpet = 10 ft × 6 ft = 60 ft²
Total area of 5 pieces = 5 × 60 ft² = 300 ft²
Area of floor = Total area of carpets
= 300 ft²
Hence, the area of the floor is 300 ft².
f) If 60 tiles each of having dimensions 2 ft. × 2 ft. are required for paving the floor of a rectangular room, find the area of the ceiling of the room.
Solution:
Area of one tile = 2 ft × 2 ft = 4 ft²
Total area of 60 tiles = 60 × 4 ft² = 240 ft²
Area of floor = Total area of tiles = 240 ft²
Area of ceiling = Area of floor = 240 ft²
Hence, the area of the ceiling is 240 ft².
g) The area of 4 walls with two windows and a door is 220 m². If each window is 3 m² and the door is 5 m², find the area of the walls excluding windows and door.
Solution:
Area of 4 walls including windows and door = 220 m²
Area of two windows = 2 × 3 m² = 6 m²
Area of door = 5 m²
Total area of windows and door = 6 m² + 5 m² = 11 m²
Area of walls excluding windows and door = 220 m² - 11 m² = 209 m²
Hence, the area of the walls excluding windows and door is 209 m².
Creative Section-A
4. Solve the following problems.a) The dimensions of a room are given in the figure aside.
[Note: No figure is provided, so I’ll assume a rectangular room with length l, breadth b, and height h. Please provide dimensions for exact answers.]
(i) Find the area of its floor.
Solution:
Area of floor = l × b
(ii) Find the area of its ceiling.
Solution:
Area of ceiling = l × b
(iii) Find the area of opposite walls along the length.
Solution:
Area of opposite walls along the length = 2lh
(iv) Find the area of opposite walls along the breadth.
Solution:
Area of opposite walls along the breadth = 2bh
(v) Find the area of its 4 walls.
Solution:
Area of 4 walls = 2h(l + b)
(vi) Find the area of the room including floor and ceiling.
Solution:
Area of room = Area of floor + Area of ceiling + Area of 4 walls
= l × b + l × b + 2h(l + b)
= 2lb + 2h(l + b)
Please provide the dimensions (l, b, h) from the figure for numerical answers.
b) A square room is 12 m long and 5 m high.
(i) Find the area of its floor and ceiling.
Solution:
For a square room, length = breadth = 12 m, height = 5 m
Area of floor = l × l = 12 m × 12 m = 144 m²
Area of ceiling = l × l = 12 m × 12 m = 144 m²
Area of floor and ceiling = 144 m² + 144 m² = 288 m²
Hence, the area of its floor and ceiling is 288 m².
(ii) Find the area of its 4 walls.
Solution:
Area of 4 walls = 2h(l + l) = 2 × 5 m (12 m + 12 m)
= 10 m × 24 m
= 240 m²
Hence, the area of its 4 walls is 240 m².
(iii) Find the area of the room including floor and ceiling.
Solution:
Area of room = Area of floor + Area of ceiling + Area of 4 walls
= 144 m² + 144 m² + 240 m²
= 528 m²
Hence, the area of the room including floor and ceiling is 528 m².
c) The length of a cubical room is 15 ft.
(i) Find the area of its floor and ceiling.
Solution:
For a cubical room, length = breadth = height = 15 ft
Area of floor = l × l = 15 ft × 15 ft = 225 ft²
Area of ceiling = l × l = 15 ft × 15 ft = 225 ft²
Area of floor and ceiling = 225 ft² + 225 ft² = 450 ft²
Hence, the area of its floor and ceiling is 450 ft².
(ii) Find the area of its 4 walls.
Solution:
Area of 4 walls = 2h(l + l) = 2 × 15 ft (15 ft + 15 ft)
= 30 ft × 30 ft
= 900 ft²
Hence, the area of its 4 walls is 900 ft².
(iii) Find the area of the room including floor and ceiling.
Solution:
Area of room = Area of floor + Area of ceiling + Area of 4 walls
= 225 ft² + 225 ft² + 900 ft²
= 1350 ft²
Hence, the area of the room including floor and ceiling is 1350 ft².
5. Solve the following problems.a) The area of four walls of a room is 210 m². If the length and breadth of the room are 12 m and 9 m respectively, find the height.
Solution:
Area of four walls = 2h(l + b)
Given: Area = 210 m², l = 12 m, b = 9 m
or, 210 = 2h(12 + 9)
or, 210 = 2h × 21
or, 210 = 42h
or, h = 210 / 42
or, h = 5 m
Hence, the height of the room is 5 m.
b) The area of four walls of a room is 639 ft². If the breadth and height of the room are 15.5 ft and 9 ft respectively, find the length.
Solution:
Area of four walls = 2h(l + b)
Given: Area = 639 ft², b = 15.5 ft, h = 9 ft
or, 639 = 2 × 9 (l + 15.5)
or, 639 = 18 (l + 15.5)
or, l + 15.5 = 639 / 18
or, l + 15.5 = 35.5
or, l = 35.5 - 15.5
or, l = 20 ft
Hence, the length of the room is 20 ft.
c) A room is 24 ft. long and 16 ft. wide. If the area of its floor and ceiling is equal to the area of four walls, find the height of the room.
Solution:
Area of floor = l × b = 24 ft × 16 ft = 384 ft²
Area of ceiling = 384 ft²
Area of floor and ceiling = 384 ft² + 384 ft² = 768 ft²
Area of four walls = 2h(l + b) = 2h(24 + 16) = 2h × 40 = 80h ft²
Given: Area of floor and ceiling = Area of four walls
768 = 80h
h = 768 / 80
h = 9.6 ft
Hence, the height of the room is 9.6 ft.
6. Solve the following problems.a) A rectangular room is twice as long as it is broad and height is 4.5 m. If the area of its 4 walls is 216 m², find the area of the floor.
Solution:
Let breadth = b m, then length = 2b m, height = 4.5 m
Area of 4 walls = 2h(l + b)
or, 216 = 2 × 4.5 (2b + b)
or, 216 = 9 × 3b
or, 216 = 27b
or, b = 216 / 27
or, b = 8 m
Length = 2b = 2 × 8 = 16 m
Area of floor = l × b = 16 m × 8 m = 128 m²
Hence, the area of the floor is 128 m².
b) The length of a rectangular hall is two times of its breadth and the breadth is two times of its height. If the area of its ceiling is 200 ft², find the area of the four walls.
Solution:
Let height = h ft, then breadth = 2h ft, length = 2 × 2h = 4h ft
Area of ceiling = l × b
or, 200 = 4h × 2h
or, 200 = 8h²
or, h² = 200 / 8
or, h² = 25
or, h = 5 ft
Breadth = 2h = 2 × 5 = 10 ft
Length = 4h = 4 × 5 = 20 ft
Area of four walls = 2h(l + b)
= 2 × 5 (20 + 10)
= 10 × 30
= 300 ft²
Hence, the area of the four walls is 300 ft².
7. Solve the following problems.a) A rectangular room is 8 m long, 6 m broad, and 4 m high. It contains two windows of size 2 m × 1.5 m each and a door of size 1 m × 4 m. Find the area of walls excluding windows and door.
Solution:
Length = 8 m, breadth = 6 m, height = 4 m
Area of 4 walls = 2h(l + b)
= 2 × 4 (8 + 6)
= 8 × 14
= 112 m²
Area of two windows = 2 × (2 m × 1.5 m) = 2 × 3 = 6 m²
Area of door = 1 m × 4 m = 4 m²
Area of walls excluding windows and door = 112 m² - 6 m² - 4 m²
= 102 m²
Hence, the area of walls excluding windows and door is 102 m².
b) A square hall is 15 m long and 5 m high. It contains three square windows each of 2 m long and two doors of size 1.5 m × 4 m. If its walls are white washed, find the area of walls covered with white washing.
Solution:
For a square hall, length = breadth = 15 m, height = 5 m
Area of 4 walls = 2h(l + l)
= 2 × 5 (15 + 15)
= 10 × 30
= 300 m²
Area of three square windows = 3 × (2 m × 2 m) = 3 × 4 = 12 m²
Area of two doors = 2 × (1.5 m × 4 m) = 2 × 6 = 12 m²
Area of walls excluding windows and doors = 300 m² - 12 m² - 12 m²
= 276 m²
Hence, the area of walls covered with white washing is 276 m².
c) Shambhu is painting the walls and ceiling of a recreation hall with length, breadth, and height 40 ft., 25 ft., and 18 ft respectively. It contains three windows of size 6 ft. × 5 ft. each and two doors of size 4 ft. × 10 ft. How much part of the walls and ceiling will be painted by him?
Solution:
Length = 40 ft, breadth = 25 ft, height = 18 ft
Area of 4 walls = 2h(l + b)
= 2 × 18 (40 + 25)
= 36 × 65
= 2340 ft²
Area of ceiling = l × b = 40 ft × 25 ft = 1000 ft²
Total area of walls and ceiling = 2340 ft² + 1000 ft² = 3340 ft²
Area of three windows = 3 × (6 ft × 5 ft) = 3 × 30 = 90 ft²
Area of two doors = 2 × (4 ft × 10 ft) = 2 × 40 = 80 ft²
Area of walls and ceiling excluding windows and doors = 3340 ft² - 90 ft² - 80 ft²
= 3170 ft²
Hence, Shambhu will paint 3170 ft² of the walls and ceiling.
Creative section-B
8. Solve the following problems.a) The length, breadth, and height of Shashwat’s classroom are 9 m, 6 m, and 4.5 m respectively. It contains two windows of size 1.7 m × 2 m each and a door of size 1.2 m × 3.5 m.
(i) Find the area of four walls excluding windows and door.
Solution:
Here, length (l) = 9 m, breadth (b) = 6 m, height (h) = 4.5 m
Area of four walls = 2h(l + b)
= 2 × 4.5 m (9 m + 6 m)
= 9 m × 15 m
= 135 m²
Area of two windows = 2 × (1.7 m × 2 m)
= 2 × 3.4 m²
= 6.8 m²
Area of door = 1.2 m × 3.5 m
= 4.2 m²
Total area of windows and door = 6.8 m² + 4.2 m² = 11 m²
Area of four walls excluding windows and door = 135 m² - 11 m²
= 124 m²
Hence, the area of four walls excluding windows and door is 124 m².
(ii) How many decorative chart papers are required to cover the walls at 2 chart papers per 8 sq. meters?
Solution:
Area of walls to be covered = 124 m²
In 8 m², 2 chart papers are required.
In 1 m², 2/8 = 0.25 chart papers are required.
In 124 m², 0.25 × 124 = 31 chart papers are required.
Hence, 31 decorative chart papers are required to cover the walls.
b) A Taekwondo hall is 30 m long, 20 m wide, and 5 m high. It contains four windows of size 2.5 m × 2 m each and two doors of size 1.5 m × 4 m each.
(i) How many player's-posters are required to hang on the walls at 3 posters per 108 sq. meters?
Solution:
Here, length (l) = 30 m, breadth (b) = 20 m, height (h) = 5 m
Area of four walls = 2h(l + b)
= 2 × 5 m (30 m + 20 m)
= 10 m × 50 m
= 500 m²
Area of four windows = 4 × (2.5 m × 2 m)
= 4 × 5 m²
= 20 m²
Area of two doors = 2 × (1.5 m × 4 m)
= 2 × 6 m²
= 12 m²
Total area of windows and doors = 20 m² + 12 m² = 32 m²
Area of walls excluding windows and doors = 500 m² - 32 m²
= 468 m²
In 108 m², 3 posters are required.
In 1 m², 3/108 = 1/36 posters are required.
In 468 m², (1/36) × 468 = 13 posters are required.
Hence, 13 player's-posters are required to hang on the walls.
(ii) How many taekwondo mats are required to pave on the floor at 5 mats per 15 sq. meters?
Solution:
Area of floor = l × b
= 30 m × 20 m
= 600 m²
In 15 m², 5 mats are required.
In 1 m², 5/15 = 1/3 mats are required.
In 600 m², (1/3) × 600 = 200 mats are required.
Hence, 200 taekwondo mats are required to pave on the floor.
Creative Section-B
9. Solve the following problems.a) The study room of Sunayana is 8 m long, 6 m wide, and 4.5 m high. It contains two windows of size 2 m × 1.5 m each and a door of size 1 m × 4 m. If she is papering the walls with 3D paper of size 4 m² each, how many pieces of wallpaper are required for her room?
Solution:
Here, length (l) = 8 m, breadth (b) = 6 m, height (h) = 4.5 m
Area of four walls = 2h(l + b)
= 2 × 4.5 m (8 m + 6 m)
= 9 m × 14 m
= 126 m²
Area of two windows = 2 × (2 m × 1.5 m) = 2 × 3 m² = 6 m²
Area of door = 1 m × 4 m = 4 m²
Total area of windows and door = 6 m² + 4 m² = 10 m²
Area of walls excluding windows and door = 126 m² - 10 m² = 116 m²
Size of each piece of 3D paper = 4 m²
Number of pieces required = Area of walls / Size of each piece
= 116 m² / 4 m²
= 29
Hence, 29 pieces of 3D paper are required for her room.
b) There is an auditorium hall of length 50 m, breadth 30 m, and height 10 m in Chhiring’s school. It contains six windows of size 3 m × 2 m each and three doors of size 2 m × 6.5 m each. If its walls and ceiling are covered with compressed fiberboard of size 5 m² each, find the number of fiberboards used in the hall.
Solution:
Here, length (l) = 50 m, breadth (b) = 30 m, height (h) = 10 m
Area of four walls = 2h(l + b)
= 2 × 10 m (50 m + 30 m)
= 20 m × 80 m
= 1600 m²
Area of ceiling = l × b = 50 m × 30 m = 1500 m²
Total area of walls and ceiling = 1600 m² + 1500 m² = 3100 m²
Area of six windows = 6 × (3 m × 2 m) = 6 × 6 m² = 36 m²
Area of three doors = 3 × (2 m × 6.5 m) = 3 × 13 m² = 39 m²
Total area of windows and doors = 36 m² + 39 m² = 75 m²
Area of walls and ceiling excluding windows and doors = 3100 m² - 75 m² = 3025 m²
Size of each fiberboard = 5 m²
Number of fiberboards = Area / Size of each fiberboard
= 3025 m² / 5 m²
= 605
Hence, 605 fiberboards are used in the hall.
10. Solve the following problems.a) The rectangular room is 20 ft long. It contains two windows of size 4 ft × 3.5 ft each and a door of size 3 ft × 8 ft. If 5 pieces of carpets each of size 10 ft × 6 ft are required for carpeting its floor and 8 pieces of paper each of size 81 sq. ft are required for pasting on its walls, find the height of the room.
Solution:
Here, length (l) = 20 ft
Area of 5 carpets = 5 × (10 ft × 6 ft) = 5 × 60 ft² = 300 ft²
Area of floor = l × b = 300 ft²
20 ft × b = 300 ft²
b = 300 / 20 = 15 ft
Area of four walls = 2h(l + b) = 2h(20 ft + 15 ft) = 2h × 35 ft = 70h ft²
Area of two windows = 2 × (4 ft × 3.5 ft) = 2 × 14 ft² = 28 ft²
Area of door = 3 ft × 8 ft = 24 ft²
Total area of windows and door = 28 ft² + 24 ft² = 52 ft²
Area of walls excluding windows and door = 70h ft² - 52 ft²
Area of 8 pieces of paper = 8 × 81 ft² = 648 ft²
Area of walls excluding windows and door = Area of paper
70h - 52 = 648
70h = 648 + 52
70h = 700
h = 700 / 70 = 10 ft
Hence, the height of the room is 10 ft.
b) A rectangular room is twice its breadth. It contains four windows of size 2 m × 1.75 m each and two doors of size 1 m × 3.5 m. If 24 pieces of tiles of size 2 m × 1.5 m each are required for tiling its floor and 41 pieces of 3D paper of size 3 m² each are required for pasting on its 4 walls, find the height of the room.
Solution:
Let breadth = b m, then length = 2b m
Area of 24 tiles = 24 × (2 m × 1.5 m) = 24 × 3 m² = 72 m²
Area of floor = l × b = 2b × b = 2b² m²
or, 2b² = 72
or, b² = 36
or, b = 6 m
Length = 2b = 2 × 6 = 12 m
Area of four walls = 2h(l + b) = 2h(12 m + 6 m) = 2h × 18 m = 36h m²
Area of four windows = 4 × (2 m × 1.75 m) = 4 × 3.5 m² = 14 m²
Area of two doors = 2 × (1 m × 3.5 m) = 2 × 3.5 m² = 7 m²
Total area of windows and doors = 14 m² + 7 m² = 21 m²
Area of walls excluding windows and doors = 36h m² - 21 m²
Area of 41 pieces of 3D paper = 41 × 3 m² = 123 m²
or, 36h - 21 = 123
or, 36h = 123 + 21
or, 36h = 144
or, h = 144 / 36 = 4 m
Hence, the height of the room is 4 m.
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